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UNIVERSITY OF 
ILLINOIS ' ‘3RARY 
URBAN < {AMPAIGN 

MATH: . "ag 


Pe 2 ee ee 


THE 
PRINCIPLES 


OF THE 


DIFFERENTIAL AND INTEGRAL 


Oo Wie Cale Ua 


APPLICATION TO GEOMETRY. 


BY WASHINGTON M’CARTNEY, Esa. 


PROFESSOR OF MATHEMATICS IN LAFAYETTE COLLEGE, EASTON, PA. 


PHILADELPHIA: 
PUBLISHED BY E. H. BUTLER & CO. 
1848. 


Entered according to an Act of Congress, in the year 1847, by 
KE. H. BUTLER & Co., 


in the Clerk’s Office of the District Court in and for the Eastern 
District of Pennsylvania. 


SMITH AND PETERS, PRINTERS. 


INTRODUCTORY NOTE. 


Tue design, in the following pages, is to exhibit the principles of 
the Differential and Integral Calculus, and to apply those princi- 
ples to Geometry. The Differential Calculus was in its origin 
geometrical, and employed by the early writers on the subject as 
an instrument for the extension of Geometry, and sciences reduci- 
ble to Geometry. Afterwards in the hands of La Grange and 
others, it became a field for analysis, and was considered to con- 
sist essentially in determining algebraically the differential coefti- 
cients of functions. In the following treatise, the principles of this 
Calculus are examined in reference to Geometry, and elucidated 
by their application to that science. 

Leibnitz, and the early writers on the Calculus, represented by 
lines the difference of two quantities where that difference was infi- 
nitely small. Subsequent analysts found fault with this plan, and 
insisted on the method of limits, as the only correct principle from 
which the mind could advance in investigations by means of the 
Calculus. In the present treatise, the language of Leibnitz has 
been retained, but (as it was indeed used by that Geometrician,) 
only for the sake of convenience. The principle of limits has be- 
come too well recognized by able writers on the subject, to be now 
departed from, and it is to be distinctly understood that in employ- 
ing in the present volume, the language and the reasoning in 
respect of infinitely small quantities, or indefinitely small quan- 
tities, convenience and clearness of geometrical conception 
have been the only motives. We have been the more in- 
duced to adopt this language, from the observation, that the wri- 
ters who studiously adhere to the language of limits in their dis- 
cussion of the principles and applications of the Calculus, often 
make the consideration of indefinitely small quantities the directrix 

lil 


810685 


IV INTRODUCTORY NOTE. 
of their reasoning. Little is therefore really gained by adhering 
rigorously either to the language of limits or to the language of in- 
definitely small quantities. No error can be introduced by em- 
ploying the latter when it is understood to be used merely for con- 
venience ; for when once a clear geometrical conception is attained, 
the transit from indefinitely small quantities to limits is a short 
journey, and readily made. We will point out in supplementar 
propositions, the mode of investigation by the principle of limits. 
The work here offered to the public, was composed from time to 

time while the author was engaged in giving instruction in the 

higher mathematics, and embodies the results of his experience in 
communicating the elements, and applications of this science, The 

Calculus is here presented as it was taught in the recitation room, 

viz. as a branch, or extension of Geometry. The author has not 

attempted, except perhaps in one or two cases, to attach any other 

than geometrical significancy to the results of Differentiation and 

Integration, Nor is it necessary to apologise here for so doing, For 

the branches of Physical science which have been subjected to the 

calculations of pure Mathematics, are those which can, by some 

arbitrary consideration, be reduced within the geometrical relations. 

For example, in Statics; forces are represented by lines, by simply 

making the lines proportional to the forces. The moment this ar- 

bitrary consideration is introduced, the relations of forces become 

the relations of lines or surfaces, and Statics is reduced to Geome- 

try. In Dynamics also, the elements of motion, viz. Time, Space, : 

and Force are represented by lines, and immediately the dynamical 

relations of Time, Space, and Force, are ascertained by ascertaining 

the geometrical relations of lines, or surfaces, This is the real 

basis of Physical Mathematics. Hence facility in the application 

of the Calculus to Geometry disposes of much of the difficulty of 

reducing Physics to the pure Mathematics, and enables the student 
to enter with ease, upon those physical investigations which de- | 
pend upon a proper conception of the geometrical relations. 

If therefore, the following pages extend the student’s knowledge 

of Geometry, and enable him to perceive, clearly and correctly, the 
spirit and mode of investigating geometrical relations by means of 


y 


INTRODUCTORY NOTE. Vv 


the Differential and Integral Calculus, the author’s aim will be at- 
tained. 

The division into distinct propositions, throughout the work, is a 
division of convenience, which has some disadvantages it is true, 
but these are more than compensated by the fact, that such a divi- 
sion directs the attention of the student definitely to a single point. 
In the applications to Geometry, the propositions are as general 
and comprehensive as consistency with clearness admitted. The 
plan has been to solve each proposition in its general form, and 
elucidate it by one or two particular cases. This plan is preserved 
throughout the Differential Calculus with the exception of a few 
propositions on the subject of consecutive surfaces, as Prop. LXI., 
LXIlI., and LXIU. If the book had been intended merely for adepts, 
a more comprehensive plan would have been adopted, but in that 
case clearness of geometrical elucidation, and consequently, adap- 
tation to the wants of the student could not have been retained. 

As no one undertakes the study of the Calculus without a com- 
petent knowledge of Algebra and Analytical Geometry, it seemed 
useless to dwell upon the algebraical details of addition, elimina- 
tion, &c. ‘To do so would increase the size of the work, without 
increasing its valuable matter, inasmuch as any student can readily 
perform such operations. The difficulty in studying most works 
on the Calculus, as well as on Analytical Geometry, arises in part 
from these very algebraic details. When presented to the eye of 
the student in the body of the work they cause him to lose sight 
of principles, and engross his attention to the exclusion of more 
valuable matters. For the same reason we have not entered much 
into the details and processes which appertain properly to books 
on Analytical Geometry. 

The theory of consecutive lines, commencing with Prop. XXXVIL., 
seemed to possess peculiar advantages in a geometrical point of 
view, and is rendered subservient to many elucidations, both in the 
Differential and Integral Calculus. The theory of consecutive sur- 
faces likewise presented geometrical advantages, and is intro- 
duced at Prop. LIX. 


The Geometry of Curve Surfaces commences at Prop. XLV., and 
|* 


v1 INTRODUCTORY NOTE, 


will be found as extensive as is usually studied in the Colleges or 
Universities in the United States, With proper preparation, the 
Geometry of Curve Surfaces presents perhaps fewer difficulties than 
are encountered in the Geometry of Curves ; especially is this the 
case, when the Geometry of Curves is first mastered. 

In the Integral Calculus the rules for integrating are as concise 
as clearness, and the nature of the subject seemed to admit. It has 
often been remarked that no branch of Mathematics requires more 
skill in algebraic processes, than the Integral Calculus. The stu- 
dent’s progress in this, is dependent altogether upon the skill which 
he possesses in algebraic operations, 

The present treatise contains the principles for integrating partial 
differential equations of the first and second orders, and higher de- 
grees of the first order. The Geometry of the Integral Calculus, 
will be found more detailed than in most books on the subject. 

Every writer, who discusses topics already investigated, must 
borrow more or less of the light of others. We acknowledge the 
advantages received from the study of Euler, La Croix, &c., but it 
is only that advantage, which accrues to the writer on any science, 
who is acquainted with what his predecessors, or cotemporaries, 
have done on the same subject 

With these remarks, the present treatise is committed to the 
public, with the hope that it may meet the approbation of Profess- 
ors, and facilitate the labors of students, in acquiring a know- 
ledge of the principles and Geometry of the Calculus. 


Easton, Pa., September Ist, 1844. 


CONTENTS. 


DIFFERENTIAL CALCULUS. 


DEFINITIONS . . \ : q i 


PROPOSITION LI. 

Given the algebraic or logarithmic equation of a curve, 
determine the ultimate ratio of the increments of its co- 
ordinates - 2 ‘ : - . 

PROPOSITION II. 

The differential coefficient deduced from the equation of a 
plane curve, represents the tangent of the angle which the 
tangent line to the curve makes with the axis of abscissas 


PROPOSITION III. 
Determine the equation of the tangent line to a given plane 
curve . . : 4 : : 
PROPOSITION IV. 
Determine the equation of the normal line to a plane curve 
PROPOSITION V. 
Determine the length of the subtangent in a given plane curve 


PROPOSITION VI. 
Determine the length of the subnormal in a given plane curve 


PROPOSITION VII. 
Determine the length of the tangent or normal of a given 
curve - - - . . - 
PROPOSITION VIII. 
Determine the point on a given curve, from which, if a tangent 
line be drawn, it will make a given angle with the axis 


of abscissas 2 fs “ 3 a 
vil 


Page 


1 


16 


26 


Vl CONTENTS. 


PROPOSITION IX. 


Determine the point on a plane curve, from which, if a tan- 
gent and normal be drawn, the subnormal will be a given 
multiple of the subtangent . - ; - 


PROPOSITION X. 
Determine the point on a curve, from which, if a tangent line 
be drawn, the subtangent. will be of a given length 


PROPOSITION XI. 
Determine the point on a curve, from which, if a tangent line 
be drawn, it will be of a given length - 


PROPOSITION XII. 
Determine the point on a curve, from which, if a tangent and 
normal be drawn, they will intercept a given distance on 
the axis of abscissas - . - . : 


PROPOSITION XIII. 
Determine the distance from a given point on a curve, toa 
given line measured on the normal to the curve : 


PROPOSITION XIV. 
Determine the point on a plane curve, from which, if a nor- 
mal be drawn, the part of it intercepted between the curve 
and a given straight line will be of a given length . 


PROPOSITION XV. 
Through a given point a tangent line is drawn, to a given 
curve, determine the co-ordinates of the point of tangency 


PROPOSITION XVI. 
Through a given point a normal is drawn to a given curve, 
determine the co-ordinates of the normal point - 


PROPOSITION XVII. 
Determine the distance from a given point to each of the co- 
ordinate axis measured on the tangent to a given curve. - 


PROPOSITION XVIII. 
Determine the distance from a given point to the tangent of a 
given curve . - é é Z . 


30 


30 


31 


33 


CONTENTS. 


PROPOSITION XIX. 


Determine the point on a curve at which the tangent is per- 


pendicular to the axis of abscissas * 


PROPOSITION XX. 
Determine when a curve has a- rectilinear asymptote, and con- 


struct the asymptote . : ; 


PROPOSITION XXI. 
Two curves intersect, determine the angle they make with 
each other at the point of intersection - . 


PROPOSITION XXII. 
Determine the relation between the parameters of two curves, 


tangent to each other - - 


PROPOSITION XXIII. 
A circle touches two given curves, determine the locus of its 


centre - - - 
PROPOSITION XXIV. 

From a given point a line is drawn, making a given angle 
with the tangent of a given curve, determine the locus of 


the intersections 5 " 


s 


PROPOSITION XXV. 
Pairs of tangents to a curve intersect ata given angle, deter+ 
mine the locus of their intersections ; - 


PROPOSITION XXXVI. 
Pairs of tangents to a curve make angles with the axis of ab- 
scissas, whose product, or sum, or difference, is constant, 
determine the locus of their intersections o* 4 


PROPOSITION XXVII. 
A circle passes through a given point and touches a given 
curve, determine the locus of its centre - - 


PROPOSITION XXVIII. 
Determine the maximum or minimum ordinate in a plane 


curve - . - - 


37 


39 


4} 


43 


49 


x CONTENTS. 


PROPOSITION XXIX. 


Given the equation of a plane curve, determine its second 
differential - - : : ; 


PROPOSITION XXX. 
If a curve be concave towards the axis of abscissas, its second 
differential is negative. If it be convex towards the axis of 
abscissas, its second differential is positive - - 


PROPOSITION XXXI. 
When a curve has a maximum ordinate, its second differential 
coefficient is negative, and when it has a minimum ordinate, 
its second differential coefficient is positive - ° 


PROPOSITION XXXII. 
Determine the differential of a circular function “ - 


PROPOSITION XXXII. 
Determine the polar subtangent of a curve . 


PROPOSITION XXXIV. 
Determine the polar subnormal of a curve - - 


PROPOSITION XXXV. 
Find the angle which the radius vector makes with the tangent 
to a curve at a given point. - . . » 


PROPOSITION XXXVI. 
Determine the locus of the intersection of the tangent and 
polar subtangent in a plane curve - - - 


PROPOSITION XXXVII. 
A line (straight or curved) is drawn, subject to a given condi- 


tion, determine the locus of the intersection of this line with - 


its consecutive line : i ° 5 2 


PROPOSITION XXXVIII. 
Determine the radius of curvature at any point in a given 
curve . ° - - - : 
PROPOSITION XXXIX. 
Given a curve, determine its evolute - - - 


Page 


59 


61 


64 


67 


73 


75 


76 


76 


78 


89 


92 


CONTENTS. 


PROPOSITION XL. 


Normals to the involute are tangents to the evolute 


PROPOSITION XLI. 
Determine the radius of curvature when the curve is referred 


to polar co-ordinates . : 3 


PROPOSITION XLII. 


- - 


Lemma : . 
PROPOSITION XLIII. 

Determine a general method of expanding into a series a 

function of the sum or difference of two independent 


variables # - : . . 
PROPOSITION XLIV. 
Determine a general method of expanding into a series a 
function of a single variable - : - - 
PROPOSITION XLV. 
Determine the equation of a plane tangent, at a given point, 
on a given curve surface : . 
PROPOSITION XLVI. 
Determine the point on a given curve surface, through which, 
if a tangent plane be drawn, its traces will make given 
angles with the co-ordinate axis - - 
PROPOSITION XLVII. 
Determine the point on a given curve surface at which the 
ordinate z is a maximum or minimum . 


PROPOSITION XLVIII. 
Determine the equation of a normal line at a given point ona 


curve surface - 
PROPOSITION XLIX. 

A tangent plane is drawn to a given curve surface, its trace 

on one of the co-ordinate planes, makes, with the co-ordi- 

nates axes in that plane, a triangle of given area, determine 


the point of tangency . : 


96 


99 


: - 101 


- 104 


- 105 


: é : - 109 


- - lll 


CONTENTS. 


PROPOSITION L. 


Page 
A tangent plane is drawn to a given curve surface, its traces 
on two of the co-ordinate planes make, with the axes, tri- 
angles, each of whose areas is given, determine the point 
of tangency : - - - : - 113 
PROPOSITION LI. 
Determine the distance from a given point on a curve surface 
to a given plane measured on the normal to the surface - 114 
PROPOSITION LII. 
Determine the distance from a given point on a curve surface 
to each of the co-ordinate planes measured on the normal to 
the surface - - . . - - 115 
PROPOSITION LIII. 
A plane is drawn through a given point and tangent toa given 
surface, determine the point of tangency . - 115 
PROPOSITION LIV. 
A plane is drawn through two given points and touches a 
given surface, determine the point of tangency - 116 
PROPOSITION LY. 
A sphere passes through a given point and touches a given 
surface, determine the locus of its centre. - ae ey 
PROPOSITION LVI. 
A sphere touches two given surfaces, determine the locus of 
its centre. : ; : . . - 118 
PROPOSITION LVII. 
Determine the differential of an implicit equation — - - 120 
PROPOSITION LVIII. 
Determine the angles which the normal line, at a given point 
on a curve surface, makes with the co-ordinate axes - 124 


PROPOSITION LIX. 
The centre of a sphere moves on a given line, determine the 


CONTENTS. 


locus of the intersection of the sphere with its consecutive 


sphere - - - 


PROPOSITION LX. 

The centre of a sphere moves along a given line, its radius 
varies as a given function of the distance of the centre 
from a given point, determine the surface which touches and 
envelopes the sphere in every position - . 


PROPOSITION LXI. 
Through a given point a plane is drawn, cutting off on the 
plane of XY a triangle of given area, determine the surface 


to which the plane is always tangent . . . 


PROPOSITION LXII. 

Through a given point on the axis of Z, a plane is drawn, 
whose traces make with the co-ordinate axes, triangles on 
the planes ZX and ZY, the sum of whose areas is given, de- 
termine the surface to which the plane is tangent : 


PROPOSITION LXIII. 

A plane is drawn, cutting off with the co-ordinate planes a 
given pyramid, determine the surface to which the plane is 
always tangent - - - - : 

PROPOSITION LXIV. 

Determine the relation between the parameters of a plane 

which passes through a given point and touches a given 


surface . . - : ; wh 


PROPOSITION LXV. 
Determine the relation between the parameters ofa plane 
tangent to two given surfaces - - . 


PROPOSITION LXVI. 
Determine the locus of the intersection of consecutive 


characteristics . - ‘ ; Z 


PROPOSITION LXVII. 


A plane touches two given curves, determine its envelope - 
+9) 


~~ 


130 


182 


133 


134 


139 


141 


ST ES 


CONTENTS. 


PROPOSITION LXVIII. 
Determine the increment of the ordinate in passing from one 
point to another on a curve surface : : - 
PROPOSITION LXIX. 
Determine the differential equations of the parallel and inclined 
sections of a curve surface - - " = 


PROPOSITION LXX. 
Determine the length of the subtangent at a given point on the 
inclined section of a curve surface . . : 


PROPOSITION LXXI. 
Determine the length of the subnormal at a given point on the 
inclined section of a curve surface : . - 


PROPOSITION LXXII. 
Determine the position of the inclined plane through a given 
point on a curve surface, when the subtangent of the inclined 
section isa minimum~— - . - - - 


PROPOSITION LXXIIL. 
Determine the position of the inclined plane through a given 
point on acurve surface, when the subnormal of the inclined 
section is amaximum—- . . . : 


PROPOSITION LXXIV. 
Determine the second differential of an inclined section of a 
curve surface : : “ : 4 - 


PROPOSITION LXXV. 
Determine the radius of curvature at a given point on an in- 
clined section of a curve surface : : . 


PROPOSITION LXXVI. 
Determine the radius of curvature of a normal section through 
a given point on a curve surface : - - 


PROPOSITION LXXVII. 
Determine the normal section whose radius of curvature is a 
maximum 5 2 . c e - 


146 


148 


149 


149 


152 


152 


153 


155 


CONTENTS. XV 


PROPOSITION LXXVIII. 


Determine the radius of curvature at a given point of an 


oblique section 2 . u _ 156 


PROPOSITION LXXIX. 
Determine the direction of a line of curvature at a given point 


on a curve surface - - - . Po iy 


PROPOSITION LXXX. 
Determine a method for eliminating an indeterminate function 160 


PROPOSITION LXXXI. 
Determine the equation of a cylindrical surface . - 164 


PROPOSITION LXXXII. 
Determine the equation of a conical surface : . 164 


PROPOSITION LXXXIII. 
Determine the equation of a tangent line to a curve of double 


curvature - - - - - - 165 
PROPOSITION LXXXIV. 
Determine the surface formed by the tangents to a curve of 
double curvature . - . . - 166 
PROPOSITION LXXXV. 
Determine the equation of the normal plane at a given point 
in a curve of double curvature - - - 167 
PROPOSITION LXXXVI. 
Determine the surface generated by the intersection of con- 
secutive normal planes to a curve of double curvature - 168 
PROPOSITION LXXXVII. 
Determine the osculating plane at a given point in a curve of 
double curvature - : - - - 168 
PROPOSITION LXXXVIII. 
Determine the radius of curvature at a given point of a curve 
: - 170 


of double curvature n 5 . 


CONTENTS. 


PROPOSITION LXXXIX. 


Determine the centre and radius 2 plana of an osculating 
sphere : - : : ; 


PROPOSITION XC. 
Determine the evolute of a curve of double curvature . 


PROPOSITION XCI. 
A tangent is drawn to a curve of double curvature, at the 
point of tangency a line is drawn, making a given angle 
with the tangent, find the locus of the intersection of this 


line with its consecutive line - - 


PROPOSITION XCII. 

A tangent line is drawn to a curve of double curvature, a right 
cone of given vertical angle has its vertex at the point re 
tangency and the tangent line for its axis, determine the 
locus of the intersection of this cone with its consecutive 
cone ‘ . : : : 2 
PROPOSITION XCIII. 

Determine the equation of a twisted surface - 


APPENDIX. 


PROPOSITION A. 
Determine the circle which touches three given curves : 


PROPOSITION B. 
A curve containing a given number of parameters touches as 
many curves as it has parameters, determine the curve’ - 


PROPOSITION C. 
Determine the equation of the cycloid . - 


PROPOSITION D. 
Determine the angle which the radius vector makes with a 


given curve at a given point - : 


PROPOSITION E. 
Determine the locus of the intersection of two consecutive lines 


Page 


171 


174 


177 


181 


185 


CONTENTS. Xvi 


PROPOSITION F. 
Page 


Determine the length of the polar subtangent of a given curve 187 


INTEGRAL CALCULUS. 


CHAPTER I. 
Principles of Integration - - . : 


CHAPTER II. 
Integration of rational fractions - - - - 197 


CHAPTER III. 


Integration of irrational fractions - : - 203 
CHAPTER IV. 
Integration of the binomial form #”~'dx (a + 62")? - 206 
CHAPTER VY. 
Integration by parts . - . - - 208 
CHAPTER VI. 
Integration of transcendental functions . - | S: 
CHAPTER VII. 
Integration by series - - - . - 217 
CHAPTER VIII. 
Integration of differentials exceeding the first order - 219 
PROPOSITION I. 
Determine the area of a plane curve . - - 220 
PROPOSITION II. 
Determine the constant added in integration - - 221 
PROPOSITION IIII. 
Determine the length of a given curve . “ - 225 
PROPOSITION IV. 
Determine the area of the surface of revolution . - 228 
PROPOSITION V. 
Determine the volume of a solid of revolution . - 230 


B 2 ¥% 


CONTENTS. 


PROPOSITION VI. 


Determine the area of a plane curve referred to polar co- 
ordinates . - . - . E 


PROPOSITION VII. 


PROPOSITION VIII. 


order, containing two variables - . : 


Determine the length of a curve referred to polar co-ordinates + 


Page 


Determine the area of a plane curve by double integration - 236 
| PROPOSITION IX. 
Determine the volume of any regular solid - - 237 
PROPOSITION X. 
Determine the area of any regular surface - - 241 
PROPOSITION XI. 
Determine the length of a curve of double curvature - 243 
. PROPOSITION XIL 
Determine the integral by means of elliptic or hyperbolic 
} arcs - - - - - : - 244 
PROPOSITION XIII. 
) Determine the integral of a differential equation of the first 
order and degree, containing two variables - - 246 
{SEOMETRICAL APPLICATIONS. ~ - - 251 
PROPOSITION XIV. 
Determine the integral of differential equations of the first 
order, exceeding the first degree : é - 263 
GEOMETRICAL APPLICATIONS. - - - 268 
PROPOSITION XV. 
‘Determine the singular solution of a differential equation ~- 271 
PROPOSITION XVI. 
Determine the integral of a differential equation of the second 
272 


CONTENTS. 


GEOMETRICAL APPLICATIONS. - - : 


PROPOSITION XVII. 
Determine the integral of differential equations of two varia- 
bles, exceeding the second order : 3 d 


PROPOSITION XVIII. 
Determine the integral of a total differential equation of three 
variables, when the first side of the equation contains one 
variable, and the other side the other two . . 


PROPOSITION XIX. 
Determine the factor which will render a differential equation 


exact - - oe - 


PROPOSITION XX. 
Determine the integral ofa differential equation of three vari- 


ables, when the variables enter it in any manner we 


PROPOSITION XXI. 
Determine the integral of a partial differential equation of the 


first order, containing three variables - . : 


PROPOSITION XXII. 
Determine the arbitrary function which enters the integral of 
a partial differential equation of the first order —- . 


GEOMETRICAL APPLICATIONS. - - - 


PROPOSITION XXIII. 
Determine the integral of a partial differential equation of the 


second order “ ° 3 F 


PROPOSITION XXIV. 
Determine the integral of a partial differential equation of the 


first order, containing four variables . - 2 


PROPOSITION XXV. 
Determine the integral of a partial differential equation of the 
first order, containing three variables, and exceeding the 


- 


first degree : : s ‘ 


280 


282 


286 


294 


300 
302 


306 


Go 
faa 
we 


314 


XX CONTENTS. 


GEOMETRICAL APPLICATIONS. - - - 


PROPOSITION XXVI. 
Determine the singular solution of a partial differential equa- 
tion of the first order - . . - 


GEOMETRICAL APPLICATIONS. - - 


MISCELLANEOUS PROBLEMS. 


PROBLEM A. 
A curve is traced in a given cylinder, determine the curve 
when the cylinder is developed on a plane . - 


PROBLEM B. 
Given the equation of a plane curve, find the curve when en- 
veloped on a given cylinder . : - ; 
PROBLEM C. 
A curve is traced on a given cone, determine the curve when 
the cone is developed on a plane - - - 
PROBLEM D. 
Given the equation of a plane curve, find the curve when en- 
veloped on a given curve surface - - - 


PROBLEM E. 
Determine the singular solution of a differential equation 


without first obtaining the complete primitive - aa 


PROBLEM F. 
Determine the area of a plane curve - 


_ Page 
32U 


co 
wy) 
bo 


333 


334 


330 


THE 


DIFFERENTIAL CALCULUS. 


DEFINITIONS. 


Ratio is the quotient that arises from dividing one quantity by 
another. ‘Thus the ratio of a to b is a ~ b. 

We will, in general, use the word line to designate either a straight 
line ora curve line. 

The first letters of the alphabet, a, b, c, &c., are in general em- 
ployed to designate constant quantities, and the last letters, w, a, &c., 
to designate variable quantities. 

The object of the Differential Calculus is to determine, by means 
of Ultimate Ratios, the properties and relations of lines, and the 
properties and relations of surfaces, and the relations of lines to 
surfaces. It is a branch, or rather a continuation of Analytical 
Geometry, and presupposes an acquaintance with the elements of 
that science. 

The nature of the Ultimate Ratios employed in the Calculus will 
be better elucidated by the first of the following Propositions, than by 
any a priori discussion. 

Equations and Quantities are usually divided into Logarithmic, 
Circular, and Algebraic. 

If we take, as an example, the equation of a plane curve, which 
involves the two variable co-ordinates and constants, we may define 
a logarithmic equation or quantity to be one that involves either or 
both of the variables in a logarithm. Thus the equation 

~¥=—alog.x+ bd 
is a logarithmic equation, the variable x being involved in a loga- 
rithm. In like manner, the following are logarithmic equations : 
y = log. (a + x), log. y = log. (a + x) 4+ 8. 


ee 
ee Sart yes Se 


A OR teen Te 


SS —s 
et TN =n Rea ee 


Zz DIFFERENTIAL CALCULUS. 


But the equations 

y = x log. a, y = (x + Db) log.a 
are not logarithmic equations ; for though they contain logarithms, 
yet the variables x and y are not involved in the logarithms. 

A circular equation or quantity, is one involving either or both of 
the variables in a trigonometrical line, or arc of a circle. Thus the 
equations 

Yf a= SIN: 2-10, = tan. (x + h) 
are circular equations, the variable x being, in each, involved ina 
trigonometrical line, 

Logarithmic and circular quantities are frequently called tran- 
scendental, as distinguished from algebraic quantities. 

Algebraic equations, and quantities, are those that do not contain 
either of the variables involved in a logarithm, trigonometrical line, 
or arcof acircle. Thus 

y = ax’? + Jb, 
are algebraic equations, and 
Yio ¢ tanoo, Uwe, ein. €, 
are also algebraic equations ; for though they contain trigonometri- 
cal lines, viz., tan. c and sin. ¢, yet c being constant, there is no 
involution of variable quantities in the trigonometrical lines. 


y= ax + 2, Poca 


PROPOSITION I. 


Given the algebraic or logarithmic equation of a curve, 
determine the Ultimate Ratio of the increments of its co- 
ordinates, 

Let the equation of the curve BP, re- 
ferred to the axes A X and A Y be 
(1) y= ar + b. 

For any point P on this curve, the 
co-ordinates w and y of equation (1) are 
A D and DP. ABD EX 

Suppose that in (1) a the abscissa of P be increased by D E, then 
putting D E = hk, and Q E = y’, equation (1) becomes 


PRINCIPLES OF DIFFERENTIATION. 3 


(2) y=a(xr+hy +, 
which is the equation of B P for the point Q. 
Expanding the second side of (2) by the Binomial Theorem, we have 
(3) y’ = ax + 8ar%h + 8axh? + ah’ + 6. 
Subtract equation (1) from (3), and we have 
(4) y'—y = Saxrzh + 8arh? + ah’. 
Divide (4) by h, and we have 


(5) Soe = 8az? + 3arh + al’ 
L 

In (4) and (5)it is evident that y'/— y= Q R, the increment of the or- 
dinate, in going from P to Q on the curve B P, and h is the incre- 
ment D E of the abscissa between the same points. ‘The first side 
of (5), therefore, expresses the ratio of the increment of the ordinate 
to the increment of the abscissa, consequently the second side of (5) 
expresses the same ratio. 

If now we suppose the increment D E, or h, of the abscissa to 
decrease, the corresponding increment Q R of the ordinate will also 
decrease, and regarding the second side of (5) as expressing the ratio 
of these increments, it is obvious that their ratio also decreases as / 
decreases. 

It is further obvious, that the ratio expressed by the second side 
of (5) cannot become less than 3az*, for that is what the second side 
of (5) becomes when h is zero. 

This value, 3a”, is called the Ultimate Ratio, or trrt of the ratio, 
because it is the limit beyond which the ratio of the increments can- 
not diminish. Equation (5), regard being had only to the second side 
of it, becomes for the Ultimate Ratio, or when h is zero, 


(6) Y—Y — 8az? 
h 
The same result may be written by observing that if the incre- 


ment h in (5) be taken indefinitely small, the terms containing / be- 
come indefinitely small, and may, when compared to the terms that 
do not contain h, be considered as zero, and consequently may be 
rejected from the equation. 

In (5), when the increment h is indefinitely small, the difference 
y'—y of the ordinates is written dy, which is called the differential 


4 DIFFERENTIAL CALCULUS. 


of y, and the increment h of the abscissa is written dx, which is 
called the differential of x. 

If therefore, we put 
(7) y'—y = dy, andh= dz, 
equation (6) may be written, 
(8) dy ata ere 

dz 

which is the Ultimate Ratio, or limit of the. ratio of the incre- 


ments of the co-ordinates of the curve whose equation is (1). 
Rigid analysis derives (8) from (5), on the supposition that in (5) the 
increment h of the abscissa decreases, and finally becomes zero, in 


which case (5) becomes (8) and ee is used as a mere symbol to denote 


: width ee ; 
the Ultimate Ratio, - being in reality ©. But inasmuch as the 
ie 0 


rules for differentiating, and the geometrical applications of Ultimate 
Ratios are more readily understood by regarding the increments of 
the ordinate and abscissa as indefinitely small, we will call these 
increments in their ultimate state, indefinitely small quantities. To 
prevent misunderstanding, it may be proper to repeat an observation 
made in the Introductory Note, that this language is adopted for the 
sake of convenience, and not for the purpose of indicating a depar- 
ture from the well established ideas in relation to Ultimate Ratios. 

The student who is familiar with the process, in Analytical Geo- 
metry, of determining the equation of a tangent line to a curve, by 
first obtaining the equation of a secant line, and then supposing the 
secant points to coincide, cannot fail to observe the similarity, and 
even identity of that process with the one just detailed. Indeed the 
Ultimate Ratio, as will be shown in the next Proposition, expresses 
the tangent of the angle which the tangent line toa plane curve 
makes with the axis of abscissas. The tangent of this angle is ob- 
tained, in Analytical Geometry, by the consideration of the secant 
line. 

But let us return to equation (8). If(8) be cleared of fractions, we 
have 
(9) dy = 8az* dz, 
which is called, the differential of equation (1). 


PRINCIPLES OF DIFFERENTIATION. 


If, instead of (1), the equation of the curve B P were 
(10) y = ax” + Db, , 
then by giving to 2 an increment h = D E, and putting y' for the 
corresponding ordinate Q E, equation (10) becomes 
(11) y =a(a+h)+. 
Expand (11) by the Binomial Theorem, and we have 
(12) y' = ax” + anz”h + an, =. gr)? +. &c., + de 
Subtract (10) from (12), divide by h, and we have 
(13) — = an” + a.n, =. 2” h + &e. 
} 
As every term after the first, on the second side of (13), contains 


h, it is obvious that when hk is made zero, or indefinitely small, the 
second side of (13) reduces to anz”-!, and then by the notation at 
(7), (13) becomes 
(14) dy moh 3 Shae 

dx 
or clearing of fractions, 
(15) dy = anz”"'dz, 
which is the differential of (10), and (14) is the differential coefh- 
cient of (10). 

In these cases, the development of (2) and of (11) is effected by the 
Binomial Theorem. In that Theorem, as is shown in most books 
on Algebra, the rules for the development are the same, whether the 
exponent of the binomial be integral or fractional, positive or nega- 
tive. Consequently we may regard the process which derived (9) from 
(1), or (15) from (10), to be as general as the Binomial Theorem, and 
we know that the development may be effected by that Theorem, 
whenever the equation of the curve is algebraic. Hence the pro- 
cess of obtaining the Ultimate Ratio in an algebraic curve is fully 
exhibited from (10) to (14). When the equation is transcendental, the 
development will be effected by other processes, as will be shown 
hereafter. We might, as is done in most books on the subject, write 
down a general form for the development of the expression, after the 
abscissa receives the increment, but perhaps a clearer idea of the 
processes can be obtained by first introducing the actual develop- 
ments, as the student is acquainted with them in his books on Alge- 
bra, Trigonometry, &c. 3 


= pn ——— — 


DIFFERENTIAL CALCULUS, 


In equation (10), if n be negative, the second term of (12), by the 
principles of the Binomial Theorem, is also negative, and (15) is nega- 
tive. Hence if we have the equation 

y= ar”, 
the differential corresponding to (15) is 
dy = — anx~"'dz. 

By comparing (9) to (3), or (15) to (12), we observe that (since 
dx =h) the second side of (9) is the second term of the development 
of the binomial (2), and the second side of (15) is the second term of 
the binomial development (12). Hence we see that the Ultimate Ratio, 
(14), viz. anx”", is the coefficient of the first power of the increment / 
of the abscissa in the development (12). Hence the Ultimate Ratio 
is called the Differential Coefficient, a term which we will hereafter 
employ as synonymous with Ultimate Ratio. 

By comparing (10) and (15), we observe that the second side of 
(15) may be produced from (10,) by multiplying (10) by the exponent 
of the variable x, decreasing the exponent by unity, and multiplying 
by dx. ‘This may be expressed by the following 


RULE I. 


To differentiate a quantity raised to a power, multiply by the 
exponent, diminish the exponent by unity, and multiply by the dif- 
ferential of the quantity. 

Thus the differential of ca* is 4 ca*dx, Here a is the quantity 
raised to the power 4, hence we multiply the quantity raised to the 
power by 4, diminish the power by 1, and multiply by dz. The 
quantity b in (10) disappears when we subtract (10) from (12), and is 
not found in (15). Hence we have 


RULE I. 


A constant added disappears in differentiating, or which is the 
same thing, the differential of a constant is zero. 
The factor a in (10) remains in the differential (15). Hence, 


In differentiating the product of a constant factor and variable 


PRINCIPLES OF DIFFERENTIATION, 


RULE III. 


the constant factor remains unchanged. 
As an application of these Rules, take the following 


x - 
EXAMPLEs. 


y= n+ ec. 


The differential of this is 


the constant factor n remains by Rule III, and the constant quan- 


dy = 5 nx‘dz, 


tity ec, which is added, disappears by Rule II. 


Differentiate the following equations : 
3 


io] 


y = axz, givesdy = Saxtdx 
y = ax* +. c, givesdy = + ax—‘dx 
y = ax—t + c, gives dy = — = ax 


If the equation were 


then regarding the exponent of x as unity, we have, by the fore- 


going Rules, 


from which we see, that if the variable rises only to the first power, 
the differentiation is effected by merely putting the characteristic d 


before the variable. 
If we have an equation of the form 


(16) y=r+t+2z+e, 


where x and z are both variables, then if when y becomes y + dy, 


y= ar + 0D, 


‘ 


dy = 1 ardz = adz, 


3 
2dx. 


x becomes x + dz, and z becomes z + dz, (16) gives 


(17) ytdy=ridr+iztdzste, 


subtract (16) from (17), and there remains for the differential of (16), 


(18) dy = dz + dz. 
Comparing (18) and (16), we have 


The differential of the sum of any number of variables is the 


RULE IV. 


sum of their differentials. 


Rule I. may be extended to a large class of binomial and poly- 
nomial expressions, as follows. 


Suppose we have 


— 


SS 


[SS 


a 


a a a Re 


8 DIFFERENTIAL CALCULUS. 


(19) y = (a + bay 

Assume 
(20) a+ bat‘ =z 
and (19) becomes 

y = 2”, this differentiated by Rule I. is 
(21) dy = nz"—"dz. 

Differentiate (20), and we have 
(22 dz = 4 ba*dz, 
substitute the values of z and dz from (20) and (22) into (21), and 
we have 
(23) dy =n (a+ bx*) ™"4ba*dz, 
which is the differential of (19). 

By comparing (19) and (23) with Rule I., we observe, that if we 
regard a + ba‘ asasingle quantity raised to the power n, Rule I. 
furnishes the direction for differentiating (19), and similar forms. 

Ex. 1.—Differentiate the equation 

y = (ax + ba*)”. 

Here we might put the part in the vinculum equal to a single 
quantity z, and proceed as was done with (19) to (23), or taking the 
part in the vinculum as a single quantity, and applying Rule I., we 
have for the differential, 

dy = n (ax + 62°)" (adx + 3 ba’dz), 
Ex, 2.—Differentiate the equations 
y = (ar—bx’), y = (az 4- ba?) i 

There is another form which may be differentiated by the imme- 
diate application of Rule I. Suppose we have 
(24) y= (2 + ay, 
in which a and z are both variable: assume 
(25) 2+r=uUu 
and (24) becomes 
(26) areal”. 

Differentiate this by Rule I., and we have 
(27) dy = nu”—"du, 


PRINCIPLES OF DIFFERENTIATION. 9 


Differentiate also (25) by Rule IV., and we have 
(28) du = dz + dz. 

The values of wu and du substituted from (25) and (28) into (27), 
we have 
(29) dy = n(x + 2)" (dz + dz). 

This is the differential of (24). 

Comparing (24) and (29) with Rule I., we observe that if we re- 
gard « + z asa single quantity raised to the power n, Rule I. fur- 
nishes the direction for differentiating (24), and similar forms. 

As an example, take the equation 

y = (ax + bz)”. 

Here we might put the part in the vinculum equal toa single 
quantity uw, and proceed as was done from (24) to (29,) or taking 
the part in the vinculum as a single quantity, we may, by Rule I. 
write down the differential, viz. 

dy = n (ax + bz)" (adx + bdz). 


Differentiate the following equations. 
1 1 


y = (a + bx)’, y = (az? + br) =, y = (ax + b)-3, 
When the exponent is 5, we may deduce from Rule I. a convenient 
rule for differentiating. For, take the equation 


(30) yi 2. 
Differentiate this by Rule I. and we have 
dx 
31 dy = te. 
(31) bag 


Comparing (30) and (31) we have 


RULE V. 
The differential of the square root is the differential of the part 


under the root divided by double the root. 
Ex.—Differentiate 


— : adz 
Pit A, gives dy = 
2/ax 
: ss 
y = (2ax—2’)’, y = (2?—z)*, 


Let us now proceed to the second part of the Proposition, viz. to 
find the differential of a Logarithmic Equation, For this purpose, let 


—— a = 


SS 


SO ee BH ee 


a 


See erent earners 


= 
— 


hater 
= r 


10 DIFFERENTIAL CALCULUS. 


(32) y = log. z 

be the equation of a curve B Q, where 
« and y are the co-ordinates of any 
point P on the curve. If a be increased 
by D E or &, then representing the cor- 
responding ordinate Q E by y’, (32 
becomes 

(33) y = log. (x + h), 

If now we could develope the second side of (33) into a series of 
monomials, we might then, by subtracting (32) from (33) developed, 
and dividing the remainder by h, obtain the ratio of the increment 
of the ordinate to the increment of the abscissa, as was done in (10) 
—(13). But as we cannot readily develope log. (x + h), the second 
side of (33), let us first subtract (32) from (33), and we have 
(34) y—y = log. (x + h)—log. a. 

Since the difference of the logarithms of two numbers is the loga- 
rithm of the quotient of the numbers, (34) becomes 
(35) y—y = log. (: ~£ ~) 

x 
and if this be developed, the result should obviously be the same as 
if we had first developed (33) and subtracted (32) from the result. 
Now as is shown in most books on Algebra, if M be the modulus of 
a system of logarithms, the log. of 1 + n is 
(38) logs (1 +n) = M (Cn ™ ak + &)), 


& 
ad 


Y 


A. Bir bD ahaa 


By making n = if (36) becomes 


. Boke ca h h? hs hi 
(37) log. (1 + eh (-- +57 ae t&) 


x 28 Ax 
This is the development of (35). Substitute it into (35), and 
dividing by h, we have 


— 1 h h? h 
38 y oe Me (Se 1 I Re oats 
( ) h x 227 .. Sx 42 ead 


This is the ratio of the increments of the co-ordinates, To obtain 


PRINCIPLES OF DIFFERENTIATION. ll 


the limit of this ratio, or the Ultimate Ratio, put h zero, or indefi- 
nitely small, and (38) becomes 


y—y ot dy ™M 
39 \ = 
( h x dx x. 
This is the Ultimate Ratio sought. In the Naperian, or hyper- 
bolic system of logarithms, M = 1 and (39) Gash 
lx 
(40) px 
x 


which is the differential of (32). 
Comparing (32) and (40), we have 


RULE VI. 
To differentiate the logarithm of a quantity, differentiate the quan- 
tity and divide by the quantity. 
This Rule gives the differential only when the logarithms are taken 
in the Naperian or hyperbolic system. For any other system of 
logarithms, we must multiply by the modulus, as is shown in (39). 


Ex.—Differentiate 


y = log. (@'+. x). 
Here a + 2 is the quantity of which the logarithm is taken ; 
hence by Rule VI. we have 


fy Pe dx 
eee 
In like manner, differentiate the equations 
: adx 
y= a log. wy Gives dy roo ey 
x 
D> 
y=alog.(x*+bzx), givesdy= “ (2adx + bdz) ed 4 
x? + ber 
y = log. (aw—a”), y = nlog. (a2) 


The n™ power of a logarithm is usually written by placing the 
after the word log. as an exponent. Thus log. “2, means the n™ 
power of the logarithm of x, and log. 2” means the logarithm of x” 
The form log. “x is the same as (log. x)". ‘To differentiate a loga- 
rithm raised to a power, we combine Rule 1. with Rule VI. 


1. >a 


sii Ds hits Ta 


12 DIFFERENTIAL CALCULUS. 


Ex.—Differentiate 


y=log."z, gives dy=n log.”""x. ~ 


y = log.” (a + 2), dy = n log.” (a + 2). a 
y = log? (ax + bz’), y = log.” (a—ba”), 

By the application of Rule VI., and the well known principles of 
logarithms, we can readily deduce Rules for differentiating a product 
a fraction, or an exponential, We will examine each of these in 
order, 

First. To differentiate the product of two variables, let us take 
the equation 
(41) = 22, 
where z and z are both variable. 

Take the logarithm of equation (41) and we have 
(42) log. y = log. + log. a. 

Differentiate this by Rule VI. and we have 
43) dy _ de, de 

Multiply this by (41) and we have 
(44) dy = xdz + zdz, 
which is the differential of (41). 

Comparing (41) and (44) we have 


RULE VII. 
To differentiate the product of two variables, multiply the second 
_ by the differential of the first, and the first by the differential of the 
second, and add the products. 
Ex.—Differentiate 
y=axzz +6, _ givesdy = axdz + azdz, 
y= avn + b,  givesdy = 2arz*dx + 3ax°x"dz, 
=zlog.z, gives dy = log, x.dz + z. cs 


The process for differentiating a product of two factors is often 
conveniently applicable when each factor contains the same variable. 


PRINCIPLES OF DIFFERENTIATION, 


Thus if we have the equation 
y = «, log. (a + =) 
we differentiate it as two factors, x being one factor, and log. (a+<) 
the other. Hence by Rule VII. the differential is 
x, dz 


a2 


dy = log. (a + x), dx + 


E'x.—Differentiate 
y = @ log. (ax + 6), y= (a + br) (m+nz), y= a" log.”2, 
1 1 


y = (x + a) (7?—1)?, ya (a—a?)? log. (a + 2) 

Proceed in the same manner for the deduction of the Rule for dif- 
ferentiating when the product is that of three or more variables. 
For if we have 
(45) i, <= Use. 
by taking the logarithms of (45), differentiating and multiplying the 
result by (45) we have 
(46) dy = uzdx + uxrdz + xzdu, 
which furnishes the Rule. 

Second. ‘To differentiate a fraction, let us take the equation 
(47) eres 

Take the logarithm of this, and we have 
(48) log. y = log, —log. z. 

Differentiate (48) by Rule VI., and multiplying the result by (47) 
we have 
(49) dy = zdx—xdz 


9 
we 
“~ 


which is the differential of (47). 
Comparing (49) and (47) we have 


RULE VIII. 

To differentiate a fraction, multiply the denominator into the dif- 
ferential of the numerator, from this subtract the numerator into the 
differential of the denominator, and divide by the square of the de- 
nominator. 

Ex.— Differentiate 
C 


DIFFERENTIAL CALCULUS. 


2 ‘ 
; az*’dzx—2axzdz 
gives dy = ; ; 
% 
; na" eda —ax"s—'dz 
gives dy = aki : 
If the numerator be constant, its differential is, by Rule I1., zero. 
Ex. Differentiate 
adx 


S 


Ee af = eee E x 
x +b x’+ log.x 

Third. An exponential is a quantity with a variable exponent. 
Thus a” is an exponential. To differentiate an exponential, let us 
take the equation 
(50) y= a*. 

Take the logarithm of this, and we have 
(51) log. y= zilogea. 

Differentiate this by Rule VI. and we have 


(52) 


y= 


Multiply this by (50), and we have 
(53) dy = log. a. a’. dx, 
which is the differential of (50). 
Comparing (50) and (53), we have 


RULE IX. 
To differentiate an exponential, multiply together the logarithm of 
the base, the exponential itself, and the differential of the exponent. 
Ex.—Differentiate 
ice ahs 
Here the exponent is 2a, hence by the Rule 
dy = log. a.a™. 2dz. 
Differentiate 
y= (a+ c)", gives dy = log. (a + c). (a+c)”nda, 
y = a®*, y = a", 
Rule IX. is deduced on the supposition that the base of the expo- 
nential is constant. If the base and exponent be both variable, we 


PRINCIPLES OF DIFFERENTIATION. 15 
can, by the process by which we passed from (50) to (53), obtain 
the differential. ‘Thus if we have 

(54) yu ul, 

where u is variable, take the logarithm of the equation, and we have 

log. y = 2 log. u 
The differential of this by Rules VI. and VII. is 
dy __ du 


== OP's, Ueithte te —. 


y u 
Multiply this by (54), and we have 


dy =) u* (10g. v dx +“), 


which shows the process. 

Differentiate 

y = (a + 2)’, y= (a+ 2’), y = (a + u")* 

The foregoing Rules are sufficient for the differentiation of alge- 
braic and ooeeithimie quantities. We might have commenced with 
the aeaueden of Rule VI. for differentiating a logarithm, and then 
have deduced Rule I. from Rule VI. For if we have the equation 
(54a) y= 2", 
by taking the logarithm of this we have 

log. y = n log. @. 


The differential of this is, by Rule VI. 


co aa 
y 2 
and multiplying this by the equation (54a), we get 
dy = "nr —'da, 


which is the differential of (54a), and gives Rule I. We preferred, 
however, introducing the Binomial Theorem to develope equations 
(1) and (10) after the abscissa had received an increment h, inas- 
much as the student is perhaps more familiar with that Theorem 
than with the Logarithmic Theorem introduced at equation (36), and 
because the development at (36) is itself deduced by the aid of the 
Binomial Theorem. 

t will be observed that Rules I. and VI. are obtained by the 
consideration that (10) and (32) are the equations of plane curves 


16 DIFFERENTIAL CALCULUS. 


The other Rules are then deduced from these two. The Curves, 


however, are merely introduced for the purpose of illustration, the 
Rules of differentiating just deduced being in fact algebraically 
true, and altogether independent of the Curves. 

There remains another class of quantities involving trigonometrical 
lines, and circular arcs. Before, however, deducing the Rules for 
differentiating these we will give a number of geometrical applica- 
tions illustrative of the use and signification of differentials and dif- | 
ferential coefficients. This will require frequent employment of the | 
foregoing Rules for differentiation. The student will carefully ob- | 
serve the difference between the differential of a quantity and the dif- 


ferential coefficient. Equation (9) is the differential of (1) and equa- 
tion (8) is the differential coefficient of (1); in like manner (15) is 
the differential of (10) and (14) the differential coefficient of (10.) 

In deducing Rules I. and VI. we gave an arbitrary increment / 
| to one of the variables x in the equation of the curve, and the cor- 
responding increment of the other variable y was deduced from this, 
and dependent upon it. This may be seen in equations (10) to (15), 
or in (1) to (9), or in (82) to (84). The variable to which the arbitrary 
increment was thus given, is called the independent variable, and 
the other is called the dependent variable. Thus in the foregoing 
deduction, x is the independent, and y the dependent variable, we 
will in general in the equation of a curve take the abscissa 2 for the 
independent and the ordinate y for the dependent variable. 

In the subsequent propositions the axes of reference are supposed 
to be rectangular except where it is otherwise specified, 


2 ge 


—— 
— 
— 
=—e 


Re me pa EE me 


on Ga 


PROPOSITION II. 


The differential coefficient deduced from the equation of 
i a plane curve represents the tangent of the angle which 
| the tangent line to the curve makes with the axis of ab- 
cissas. 


=a SS 
SP ee 


METHOD OF TANGENTS. 17 


FIG. 3. 


Let the equation of the curve 
H P be 


(55) y=ar'+b, zc 
At any point P whose co- 
et 


ordinates are x and y suppose a G@\ + 
This tangent line may be ae A DE x 
sidered as coinciding with the curve along the indefinitely small 
distance PQ. The point Q being taken indefinitely near to P, the 
line P L, which is parallel to the axis A X, represents the increment 
of the abcissa 2, and Q L the increment of the ordinate y. P L and 
€ L may therefore be taken as the representatives of da and dy. 
With the centre R and radius R K=1 describe the arc K G and 
draw K V tangent to the ark at K. 
By similar triangles R K V and P L Q we have 
Gales KoVas seals shah 


or since K V is the tangent of R, 


tangent line R P be drawn. 


(56) 1: tan.R:: dz: dy 
from which we have 
(57) eh te: 

dx 


But the angle at R is the angle which the tangent line, tangent at 
the point P, or (x, y), makes with the axis of abscissas. Hence (57), 
which shows that the tangent of this angle equals the differential co- 
efficient, proves the proposition. 

Equation (57) is independent of the equation (55), not being deduced 
from it, consequently the proposition is true whatever be the nature 
of the plane curve H P. ‘To obtain the tangent of the angle R for 
any given curve (55) we may therefore deduce the differential coeffi- 
cient from (55) and its value is the tangent of the angle required. 

As a particular example the differential coefficient from (55)is by 
Rule I. 

(58) dy _ 3 aa 
dx 
Consequently for the angle R we have in this particular case 
t 


' i 


18 DIFFERENTIAL CALCULUS. 


(59) tan. R=3 aa’, 

In 59 the tangent of R depends upon the value of the abscissa x. 

The Rules of Proposition I. are algebraically true, and explain the 
mode of differentiating algebraic and logarithmic equations; and when 
differentiated, the differential coefficient may be deduced by the ordi- 
nary rules of Algebra. Proposition I. shows the point of connexion 
between these algebraic Rules and Geometry, viz: ‘The differen- 
tial coefficient represents the tangent of the angle,” &c. From this 
point of connexion a great system of Geometry may be constructed, 
inasmuch as almost EVERY GEOMETRICAL INVESTIGATION, BOTH OF 
LINES AND OF SURFACES, INVOLVES, EITHER DIRECTLY OR INDI- 
RECTLY, THE TANGENT OF THIS ANGLE. ‘The following Proposi- 
tions, from page 20 to page 59, exhibit the mode of resolving 
problems in plane Geometry, by aid of the principle established in 
this Proposition I], At Proposition XLV. commences the application 
of the same principle to the solution of problems in the Geometry of 
surfaces. ‘The pure theory, or Rules of differentiation, are exhibited 
in Propositions J., XXIX., and XXXII., which (if it is preferred) 
may be first studied; and then the application to Geometry of the 
principle evolved in Proposition I]. may be perused in the other 
Propositions, as far as time and circumstances admit. 

To obtain the differential coefficient from an implicit equation 
such as a? y? + b? a? —a? b? = 0, we may either first solve the 
equation for y and then differentiate, or (which is more convenient 
in practice) we may first differentiate each term of the equation, and 
then deduce the differential coefficient by the ordinary rules of algebra. 

Before proceeding farther, it may be advantageous to explain a 
notation that has been found very convenient. If we have any 
number of curves whose equations are 
(AQ gle a? ep) Sy = ar a8, ay (RR? Sat, 
where y equals an expression into which x enters we may represent 
them all by the single form. 

(603 y = 02, 
which is read, y equals a function of x, and signifies that y equals 


METHOD OF TANGENTS. 19 


an expression composed of a and constants. Equation (60) may 
therefore, be used to designate dny curve whose equation is solved 
for y, and is a general form embracing all such equations as those 
marked (A.) 

Instead of g in (60) any other character may be employed for 
the same purpose, ‘Thus, 

ya! Ba; y= Vx, yofe yor, 

would each mean the same as (60) viz. that y equals an expression 
composed of x and constants. 

If the equations of the curves instead of being solved for y as in 
(A) were of the form, 


(B) lived + Bae R? — 0» y? — px =O, 

\ ry—a’=o0, ax —ry ry — ob =n0, 

i. e. not solved for y, then they may be represented by the form, 
(61) Q (x,y) = 9; 


which is read, a function of xy equals zero, and means that the 
equation of the curve contains the co-ordinates x y and constants, as 
in any one of the group B. 

Equations of the form A or (60), i.e. solved for y are said to 
contain y as an explicit function of 2. 

Equations of the form B or (61), 1. e. not solved for y are said to 
contain y as an implicit function of x. One quantity is said to be 
a function of another when it depends upon it. ‘Thus in the curve 
H P, the ordinate P D or y is said to be a function of the abscissa 
A D ora, because if the abscissa be varied the ordinate will also 
vary, or in equation (60) which is the explicit form of the equation 
of any plane curve whatever, if x be increased or diminished y will 
also vary, being connected with # in the equation. Hither of the 
forms (60) or (61) may be used to designate any plane curve what- 
ever, (60) being used when the equation is explicit,and (61) when it 
is implicit. 

In the case of the parabola, (60) is y = J px, the explicit form, 
and (61) is y* — px = 0, the implicit form, 

In many of the following propositions we will use the explicit 
form (60), of the equation of a plane curve, employing it as a mere 


aul — 


SSS 
Somme 
5 


20 DIFFERENTIAL CALCULUS. 


symbol to denote the equation of any curve. We might for the same 
purpose, and in the same manner, employ the ioiogs form (61). 
We may arrive at the same result (57) by the consideration of 
limits, as follows: Let (x,y) be the co-ordinates of the point P 
(fig. 2.) and (a',y') of the point Q, and through the points Q and 
P draw the secant line Q R, it is obvious that y —y = Q Land 


« — x= PL, and the two triangles R K V andPQL being, as 
before, similar, we have as in (56). 

(56a) 1:tan,.R::,2'—a:y'—y, 

from which we have 

(57a) tan. R = slap 


which is the tangent of the angle which the secant line Q R makes 
with the axis of a. This angle varies the nearer the point Q ap- 
proaches to the point P, and in the limit, when Q coincides with P 
the secant line Q R becomes a tangent line P R, the angle at R be- 
comes the angle which the tangent line makes with the axis of 2, 
d 
and sae becomes “n the ultimate ratio or differential coeffi- 
x —2 ne 
cient, Hence, when P R is a tangent line, (57a) becomes (57). 


PROPOSITION, III. 


Determine the equation of the tangent line toa given 
plane curve. 


FIG, 4. 


Let the equation of the plane 
curve H P be represented by 
(62) y = gee 

Let x’ and y’ be the variable 
co-ordinates of the line R P and x 
« and y the co-ordinates of the R oi D G 
point of tangency P, The equation of any line R P passing through 
the point P is, 

(63) y —y=—a(r' —2). 

Since a in (638) is the tangent of the angle P R X, we have by 

(57), when P R is a tangent line, 


METHOD OF TANGENTS, 21 


dy 
(64 ae = (he 
(64) dx 
Put this value of a into (63), and we have 
(65) yy = dy (z'— <=), 
dx 


which is the equation of the tangent line P R. This is a general 
form, being independent of the nature of the curve H P. To apply 
it to any particular curve, we deduce from (62) the value of the dif- 
ferential coefficient, which being put into (65), gives the tangent line 
to the particular curve. 

As an example, let (62) be for a particular curve, 


(c) y= art b, 
From this we get 
(d) YY — sax%, 
dx 


This value of the differential coefficient put into (65,) we have for 
the tangent line to the curve(c),the equation 
(e) y'—y = 8aa? (x'—2). 

In the general form (65), and in every particular case, it is to 
be observed that 2 and y are the co-ordinates of the point of tan- 
gency. By means of the equation of the curve (62), we can elimi- 
nate y from the tangent line (65). 

If we designate the differential of ox, by writing the character- 
istic d before it, thus, d. oz, we may write the differential coeffi- 
cient of (62) in the form 


= di d. Ov 
(65a) = os ‘fn 
or putting 
d. ox ; 
= Wa 
dx ‘zs 
we may write (65a) 
(65d) e = 92, 


and by substituting from (62) and (655) into (65), we have for the 
veneral form of the tangent line toa plane curve, 
4% 


22 DIFFERENTIAL CALCULUS. 


(65c) y—opr = 9'2 (a' —x). 

But it is simpler to employ (65). 

From (658) we observe that when the curve (62) is explicit, the 
differential coefficient is a function of x. 

If the equation of the curve be implicit, as in form (61), the pro- 
cess is the same, viz., deduce from (61) the differential coefficient, 
and put its value into (65). 

Eix.—Determine the equation of the tangent line to the curves 

yar +e -. y'—y = 4aa’ (x'—2), 
y =ar—a. *.  y'—y = (a—2z) (a'—z) 
(C) Se aaa ene (x + h), Visamilog 2, 


xv + y?—R? = 0, at “ha8 abt — = 0, y—pr =o, 


PROPOSITION IV. 


Determine the equation of the Normal line to a plane 


curve. 
FIG. 5, 


Let PG be the normal line P 
to the curve at P, and (2’,y') the 
co-ordinates of any point on 
PG, and (x,y) the co-ordinates 


of the normal point P. R “A. Dp Ge 
The equation of any line P G through P is of the form 
(66) y—y = a (x tS 


where a is the tangent of the angle PGX. 
If PR be tangent to the curve at P, and (65) its equation, we 
have by the condition of perpendicular lines, (Analytical Geometry.) 


dy _ 1 
ae a” 
or taking the reciprocal, 
dz 
OS —p 
dy 
This value of a put into (66), we have 
dx 


(87) Y—y ie: (7— x), 


METHOD OF TANGENTS. 23 


which is the equation of the normal line P G. If (62) be the equa- 
tion of the curve H P, we deduce from (62) the differential coeffi- 
cient and put its reciprocal into (67). 

Ex.—Determine the equation of the normal line when the curve is 


dy 
= 03 a Sax’, 
(c) y=—ar-+b oe 
and the reciprocal of this put into (67), we have 
(d) y—y — __ t (x'—2), 
Sax’ 


the normal line required. 
Determine the normal line in the curve 
(e) e’t+y—R=o, 
From this we have 


and (67) becomes 
(f) yy = “ (x'—a), 


which is the equation of the faint line to the curve (e). 
Ex.—Determine the normal line in the curves marked (C), 


PROPOSITION V. 


Determine the length of the subtangent in a given plane 
curve. 


FIG. 6. 
Let the equation of the given r—c¢ 
curve H P be 
(68) Y = or, 
where x and y are the co-ordi- x 
nates of the point of tangency R J.4 D . 


P. The subtangent is RD, To find the length of RD we have, 
by Trigonometry, i in the right-angled triangle P R D, 
(68a) PD=RD tan. R. 


SS 


ewe 


~— 


See 


iabpaipenigoationdamncnath 


24 DIFFERENTIAL CALCULUS. 


Put into this the value of tan. R at (57), and we have, since P D 
is the y of the point P, 
(69) rina ng fal 

dy 

which is the length of the subtangent required. To apply (69) to 
any given curve (68), deduce from (68) the differential coefficient, 
and its reciprocal put into (69), we have the subtangent for the par- 
ticular curve. 

E.x.—Let the curve (68) be 


et) tim So Seen aoe 2ax, or ssid ae oe 
dx dy 2ax 
This put into (69), we have 
RD=_4 _ 


2ax 
the subtangent required. 
Ex.—Determine the subtangent in the curves (C), 


PROPOSITION VI. 


Determine the length of the subnormal in a given plane 

curve. 
FIG, 7 

Let the equation of the given 2 
curve H P be | 
(70) y= om. 

Draw P G the normal, then 
D G is the subnormal whose#Z ry nee 
length is required. 

The right-angled triangles R P G and D P G, being similar, we 
have the angle P R D equal to the angle D P G, Hence by (57) 


(71) YY _ tan. D PG. 
dx 


From the right-angled triangle P D G we have, by Trigonometry, 
*(72) D "Gea 5P Ditan. DP Ge. 
By (71) this becomes, (since P D is the ordinate y), 


to 


METHOD OF TANGENTS, 


(73) DG 9 wD subnormal, 
C 


We might eliminate y from (73) by means of the equation of the 
curve (70), and the subnormal wonld be a function of a, 

To apply (78) to any given curve, deduce from the equation of 
the curve (70), the differential co-efficient, and put its value into (73). 

Ex.—Determine the subnormal of the curve 


(c) Y Ge +L 0. a OE, 


and (73) becomes 
(d) DG = 3az2’ y, the subnormal. 
[f we eliminate one of the co-ordinates, as y from (d) by means of 
(c), we have for the subnormal as a function of 2, 
(e) DG = 38ax? (ax* + 5). 
Kix.—Determine the subnormal in equations (C). 


> 


PROPOSITION VII. 
Determine the length of the tangent or normal ofa given 
curve. 
Let HP [Fig. Proposition VJ.] be the given curve, and (70) its 
equation. PR isthe length of the tangent, and PG of the normal. 
From the right-angled triangle PRD we have 


aaa ey: = 
PR = PD°+-RD. 
Put in this for RD its value at (69), and since PD is the ordiaate 
y, we have for the length of the tangent 


2,2 
(74) PR=(¥ + # S), 
In like manner, for the length of the normal PG we have 
pee | es '- = 
PG =3PD + DG, 
Put in this for DG its value at (73), and we have for the length 
of the normal 


(75) PG (y gee 


26 DIFFERENTIAL CALCULUS. 


The differential coefficient in (75) and (74) is to be supplied from | 
(70), the equation of the curve. | 
Ex,—Determine the length of the tangent and normal lines in | 
equations (C.) | 


PROPOSITION VIII. 


Determine the point ona given curve, from which, if a : 
Wi tangent line be drawn, it will make a given angle with the 
axis of abscissas. 


FIG, &. 


Let the equation of the given 


curve HP be 
(76) y = oe. 
Since the angle PRX, which 
q/| the tangent line makes with? BINS D 


axis of x is given, its tangent is given. Hence putting m for the 
tangent of PRX, we have by (57) 


‘ha wey d 
i {aK ( 4 7) = SS if 

The two equations (76) and (77) solved for a and y make known 
the co-ordinates of P, the point of tangency required. 

Ex.—For a particular curve let (76) be 


c Le 0 or dy = 2azx and (77) becomes 
( y d 
Le 


ee 


Hh (2)  2ax = m, equations(c)and(d)solved for x and y give 


(€) oS Le ee 
Aa 


which are the co-ordinates of P the point required. 
Ex.—Determine such a point of tangency in equations (C), 


PROPOSITION IX. 


Determine the point on a plane curve from which if a | 
tangent and normal be drawn, the subnormal will be a . 
given multiple of the subtangent. 


METHOD OF TANGENTS. 27 


FIG. 9 


Let the equation of the given P=<o¢ 
curve HP be 
(78) y = 9a. 

Let m be the multiple that i. 
DG is of RD, then by the3 ort D G 
proposition 
(79) DG = m. DR 


substitute into this for DG and DR, their values in (73) and (69), 
and we have 


(80) y dy —— ie y dx 
dx dy 
from which we have 
dx — 


Equations (78) and (81) are sufficient to determine x and y, the 
co-ordinates of the point required, 
Ex.—For a particular curve let (78) be 
(c) y = ax? + b, From this we get 
dy _ 


and (81) becomes 
(d) 2ax2-—= + /m, 
solve equations (c) and (d) for 2 and y and we have the co-ordin- 
ates of the point required. 
Ex,—Determine such a point in curves (C), 


PROPOSITION X. 


Determine the point on a curve from which if a tangent 
line be drawn the subtangent will be of a given length. 
Let the equation of the given curve HP (fig. 9.) be 


(82) y= ur, 

putting n for RD the given subtangent, (69) gives 
an dx 

(83) Ria, Fs, 


b 
# 


Mi 
f 
¥ Ve 
vb 
' 

P s 
\ 
f 
" | 
ai 


ae = 


= 


PA ee ee SE en 
(ge an 


a 


| 
j 
ti 
ii 

I 

My 
My j 
j 1) 
ti 

\& 

i | 


28 DIFFERENTIAL CALCULUS. 


after putting into (83) the value of “ 
y 


deduced from (82), equations (82) and (88) solved for « and y make 
known the point required. 
Ex.— Determine such a point in curves (C). 


PROPOSITION XI. 


Determine the point on a curve from which if a tangent 
line be drawn it will be of a given length. 
Proposition X. suggests the method of doing this. 


PROPOSITION XII. 


Determine the point on a curve from which if a tangent 
and normal be drawn they will intercept a given distance 
on the axis of abscissas. 


Let the given curve HP be 
represented by 
(84) y = Ou. 

It is obvious that RG the 
part of the axis of abscissas in-_ 
tercepted by the tangent PRE wa D G 
and the normal PG is composed of the subtangent and subnormal. 
Hence, putting c for the given intercept RG we have by means of 
(69) and (73) 

(85) eae a gi 

Having put into (85) for the differential coefficients, their values de- 
duced from (84), the equations (84) and (85) make known the co- 
ordinates x and y of the point required. 

Ex.—Determine such a point in curves (C). 


METHOD OF TANGENTS. 29 


PROPOSITION XIII. 


Determine the distance from a given point on a curve to a 


given line, measured on the normal to the curve. 
; FIG. 11. 
Let (x, y) be the givenco-ordin- py HH 


ates of the point P, and let the py 
equation of the curve HC be 


and the equation of the given line a. 
: N 
MN be fe = 
(87) y —azr' + Bb. 
The equation of the normal PQ is by (67’) 

(38) oP ep hae) 
where for brevity we put 

., dy 

~ dx 


and where 2’, y' are the variable co-ordinates of PQ. ; 
If x and y be the co-ordinates of P, and x’ and y’ of L, the dis- 
tance PL, which call N, is by a well known form in Analytica! 
Geometry, 
(39) N? = (y' — yf + (2 — ay’, 
At the point L, x’ and y’ are common to (89), (87) and (88). 
Find the values of 2’ and y’ from (87) and (88) and substituting 
them into (89) we have for the length required, 


nie 


(90) en Fe as 
ap-+l1 
The differential coefficient p deduced from (86) and its value put 
into (90), we have the distance PL in terms of the given co-ordin- 
ates of P and other known quantities. 
Determine such a distance for the following curves. 
(D) y= az’, P+ yY—R=0, @ytibPwe—avb’ =o, 
D 2) 


DIFFERENTIAL CALCULUS. 


PROPOSITION XIV. 


Determine the point on a plane curve from which if a 
normal line be drawn the part of it intercepted between the 
curve and a given straight line will be of a given length. 

This is obviously effected by solving (86) and (90) for x and y, 
the co-ordinates of the point P. Hence, putting for the differential 
coefficient p in (90) its value from (86), we have (86) and (90) to 
find « and y in terms of the given line N and other known quantities, 

Propositions similar to XIII. and XIV, might also be solved in re- 
lation to the tangent line. The method of the solution is readily 
suggested by these Propositions. 


PROPOSITION XV. 


From a given point a tangent line is drawn to a given 


curve, determine the co-ordinates of the point of tangency. 
FIG. 12. 


x 


Let M be the given point, and (m, 7) 
its co-ordinates. ns 
Let the equation of the given curve 


PD be 
A. 


(91) o(xr,y) = 0 E x 
Let MP be the tangent line. From (65) the equation of MP is 
(92) y —y = p(t —2) 


where p is put for the differential coefficient, an abbreviation we will 
frequently employ hereafter, and «' and y’ are the variable co-ordin- 
ates of MP. 

At the point M, 2' and y’ become m and n, and (92) becomes 
(93) 2—y=p(m— 2). 

For the differential coefficient p in (93) put its value taken from 
(91), and then solving (91) and (93) for x and y we have the co- 
ordinates of the point P required. 

Ex. 1. Let (91) be the circle, 

(a) a+ y—R = o. 
The differential coefficient from this is 


METHOD OF TANGENTS, 31 


(bd) dy ee P, 
dx y 
and (93) becomes 
(c) m— Y= ft. (m — 2), 
y 


solve (a) and (c) for x and y, and we have the point P required. 

Ex. 2. Determine such a point of tangency in the curves (D.) 

[t is obvious that if (91) be a curve of the n‘ degree, there will 
be n points of tangency, all of which will be made known by the 
n values of 2 and y found by solving (91) and (93). Thus, example 
(a) is of the 2* degree, and the solution of equations (a) and (c) 
gives two values for x and two for y, which shows, that from the same 
point M two tangent lines can be drawn to the circle (a). 


PROPOSITION XVI. 


From a given point a normal is drawn to a given curve, 
determine the co-ordinates of the normal point. 

By the normal point, a term already used, is understood the point 
on the curve where the normal intersects it. 

From (67) we have for the equation of the normal line 
(93a) y—yr Se’ — r). 

Pp 

Proceed with this equation as with (92), and the solution is the 
same as in Proposition XV. 

Ex.—Find such a point in the curve 


(b) a the, 
From this, p = 2ax, and (93a) becomes, (putting n and m for 
y' and 2’) 
1 , 
(c) a— Y= — (m — 2), 
2ax 


: y “ 
Proceeding with the solution of (b) and (c), we find a cubic equa- 
tion which shows that from any point not on the curve, three nor- 
mals can be drawn to the curve (5). 
Ex.—Determine such a point in the curves (D). 


= ae Se ee 
~ 


Saas 


a Nn cae 


ee ee 


32 DIFFERENTIAL CALCULUS. 


PROPOSITION XVII. 


Determine the distance from a given point to each of the 


co-ordinate axes, measured on the tangent to a given curve. 
FIG. 13. 


Let M be the given point, and 
(m,n) its co-ordinates. Let the 
equation of PD, the given curve, 
be represented by 
(94) P(ty) = 0. E 

Through M suppose the tangent line PC ae Its equation is, 
as in (98), for the point M, 


(95) N—Yy = p(m—z). 
The general equation of the same line is, as in (92), 
(96) y—y = p (v’—2). 


To get the distances AH and AC, put first in (96)y' zero, and 
we have for 2’, 


(97) Dace eh a } =— AC, 


Again, put in (96) 2’ zero, and we have for y’, 
(98) y =y—px= AH. 

Fiquations (97) and (98) make known AC and AH in terms of 
x,y, and constants, (@ and y being a co- prone of the point 
of tangency P.) }— HY 

Having also the co-ordinates (ithe) of the point M, it is evident 
that the required distances MH and MC are readily expressed from 
the right-angled triangles CEM and HSM, in terms of AH, AC, AE, 


and EM. 
These expressions for MH and MC will involve the co-ordinates 


«and y of the point of tangency, which co-ordinates may be de- 
termined by solving (94) and (95) for x and y. 

Determine these distances in the curves (D). 

If the proposition were to determine similar distances measured 
on the normal line to a given curve, the solution would be effected 
in the same manner, by taking the equation of the normal line in- 
stead of (95) and (96). 


METHOD OF TANGENTS. 33 


PROPOSITION XVIII. 


Determine the distance from a given point to the tangent 
of a given curve. 


FIG. 14, 


Let PD be the givencurve, and y 
(99) ?(2,y) =='D, 
its equation, 

Let (m,n) be the given point M, 
and 2’ and y’ being the variable co- © 
ordinates of PS, the equation of the tangent line PS is 
(100) y—y = p (#'—2). 

From M draw MH perpendicular to PS, and MH is the distance 
required. For the equation of the line MH, passing through M and 
perpendicular to PS, we have 


(101) y—nr = — a (7'—m),. 
B 


For the required distance MH we have, by Analytical Geometry, 


(102) MH = (y'—n)? + (2'—m)’, 
where x’ and y’ are the co-ordinates of the point H. 

At the point H, x’ and y’ are common to (100), (101) and (102), 
hence solve (100) and (101) for x' and y’, and putting their values 
into (102), the required distance MH becomes known. This dis- 
tance is expressed in terms of constants, and of the co-ordinates x 
and yof the point of tangency. ‘These co-ordinates are given by 
the proposition. 

Determine this distance in the curves (D). 


PROPOSITION, XIX. 


Determine the point on a curve at which the tangent 
line is perpendicular to the axis of abscissas. 


5 * 


" a a gee a = oe 
a 


3e4 DIFFERENTIAL CALCULUS. 


Let PD be the given curve, and y EG (36 
(108) oxy) = 0, 
its equation. 

Let P be the point at which the tangent 
line PE is perpendicular to the axis of 0 


abscissas. 
Since © expresses the tangent of the angle which the tangent 
wv 


lines makes within the axis of abscissas, we have, when this is a right 
angle, 

ZY _ Infinity, 

dx 
or since the reciprocal of infinity is zero, 
(104) na 
dy 

Equations (103) and (104) serve to determine x and y, the co- 
ordinates of the point required. 

The same result would be obtained by considering that the co- 
tangent of an angle is the reciprocal of the tangent, and that when 
the angle E is aright angle, its co-tangent is zero, which gives (104). 

Ex.—Determine such a point when the curve PD is the circle. 


0. 


(0) (z—a)? + (y—s)' — R? = 0, 
Differentiate this, and we have 
(c) 2 (x—a) dx + 2 (y—B8) dy = o. 
From this we get 
(d) Cgc atte aie 
dy r—a 
By (104) this is zero. Hence 
U het ae 
(e) Tine con ok 0. 


Solve (e) and (6) for x and y, and we have 
= 6, c=a ot R, 

which shows that there are two points on the circle (b) at which the 
tangent line is perpendicular to the axis of abscissas. 


ASYMPTOTE TO A CURVE. or) 


Ex.—-Determine such a point in the curves 
(E) y = ax’ + J, (y—s) —nz = 0, y? = me. 


PROPOSITION XX. 
Determine when a curve has a rectilinear asymptote, and 


construct the asymptote. 
Definition.—A rectilinear asymptote is a tangent line whose point 


of tangency is infinitely distant. 
Such a line consequently approaches continually nearer to a 


curve without, however, touching it at any finite distance. 
FIG. 16. 


Mie 


Let PD be a given curve, and x. 
(105) y = ox 
its equation. The equation of a 
line tangent to this curve is of the 


pet 


form a 
(106) y—y = p (r¥—2), 
where x and y are the co-ordinates of the point of tangency. In 
(106) put y’ = o and we have 


| p 
Again in (106) put z' = 0, and we have 
(108) y = y— pz.= AH, 


The differential coefficient p derived from (105) obviously does 
not contain y. It may therefore be regarded as a function of 2. 
Let us then for brevity put 


and substitute (105) and (109) into (107) and (108), and we have 
(110) ye Seer aD al 
Ya 
and 
(111) AH = or — xyz. 


In order to determine whether PD can have an asymptote, we 
make z infinite in (110) and (111), and observe what AC and AH 


ating 


SS 


== >= ss 


a 


SN eee 


SS 
— 


36 DIFFERENTIAL CALCULUS. 


then become. If both the values of AC and AH become infinite 
when « is infinite, the curve has no asymptote, for no line can then 
be drawn which will become a tangent at a point infinitely distant. 

ff either or both of the values of AC and AH become finite or 
zero when 2 is infinite, the curve has an asymptote, because a line 
drawn in a definite manner becomes a tangent ata point infinitely 
distant. 

If both the values of AC and AH are definite lines when 2 is in- 
finite, the line drawn through the points C and H is the asymptote. 

If both these values are zero for x infinite, the asymptote passes 
through the origin A, and to draw it, we must know the angle it 
makes with one of the co-ordinate axes. 

The tangent of the angle which the asymptote makes with the 
axis of x is found by making 2 infinite in (109). 

If for x infinite one of the intercepts (110), (111) is infinite, and 
the other finite, the curve has an asymptote parallel to the axis on 
which the intercept is infinite. If one of the intercepts (110,) (111) 
be infinite, and the other zero, the axis is the asymptote. 

If for any given value of « in (110), (111), one of the intercepts 
AC, AH becomes infinite, and the other finite or zero, that value of 
x is the abscissa of a point which is infinitely distant, and the curve 
has an asymptote parallel to the axis of y, if a be an actual value, 
and coinciding with the axis of y, if x be zero, 

We will illustrate this by an example. 


Take as a particular case of (105) the equation 
: eitasaas b ei Ae a 
(¢) a (z"—a ys 
which is the equation of the hyperbola. 

Proceed with the equation of its tangent line as above directed, 
and we find the intercepts corresponding to (110), (111), each zero 
when 2 is infinite. Hence the hyperbola (c) has an asymptote pas- 
sing through the origin. 

To draw this asymptote we must find the angle it makes with the 
axis of 2. For this purpose, the differential coefficient deduced from 
(c) may be put into the form 


ANGLE OF TWO CURVES, 37 


(d) Pp pag =G-5)- 


which corresponds to (109), When = is infinite, (d) becomes 
(e) p=b~a, 
which makes known the angle that the asymptote makes with the 
axis of 2; and the asymptote, passing also through the origin, can 
be drawn, 

Ex. 2,—Determine whether the curve 


(f) y= a 

has an asymptote. Differentiate this, and we have 
dy a’ 

(g) —— == —— — 
dx = 


This is the value of Ja in (109). Hence substituting from (/) 
and (g) into (110) and (111), we have 
(h) AC = 2z, and AH = 


2a" 


2 

If in the values (h) x be made infinite, we have AC infinite, and 
AH zero, Hence the axis of x is an asymptote. If in (h) x be 
zero, AC is zero and AH is infinite, from which we infer that the 
axis of y is also an asymptote. 


PROPOSITION XXI. 


Two curves intersect, determine the angle they make with 
each other at the point of intersection. 


Let HD and ES be the 
intersecting curves, intersect- ¥Y 
ing at the point P. 

Let the equation of ES be 
represented by 
(112) 9(a,y) =0, 
and the equation of HD by 
113) 1(x,Y) = 0, 


38 DIFFERENTIAL CALCULUS. 


‘The angle made by two intersecting curves is the same as the 
angle made by their tangents at the point of intersection. At the 
point of intersection P, draw PN and PL tangent to the curves HD 
and ES respectively. Let 

p = differential coefficient of (112) = tan. PLX, and 
p = differential coefficient of (113) = tan, PNX. 

If V = tangent of the angle NPL, then since NPL is the differ- 
ence of the angles PLX and PNX, we have, by a well known Theo- 
rem in Trigonometry, for the tangent of the difference of two 
angles, 

(114) WV hea SEED 
1 + pp 

This is the value of the required angle in terms of the differential 
coefficients p and p’. As pand p’ are functions of a and y,if the 
values of p and p' be deduced from (112) and (113), and substi- 
tuted into (114), V will then be a function of 2 and y, and we may 
represent (114) by 
(114a) Viren) 
where F (ay) is put for what (114) becomes when the values of p 
and p' are substituted into it. Now solve (112) and (118) for x and 
y, the co-ordinates of the point of intersection, and their values put 
into (114a), we have the tangent of the required angle in known 
terms. 

Ex.—Find the angle of intersection of the circle and parabola. 

Here (112) and (113) become 


(c) x? + y*? — R*® = o, and y? —4mx = 0, 
from which we get respectively, 

dy x dy 2m 
d Ne lente aaah ee SA 2h Ssh) Eee 
@) dx y : dx y p 

The values of p and p’ in (d) put into (114), we have for (114a), 
(€) Vv= Ai 2m) o* 
2mx — 1 


Solve equations (c) for x and y, and putting their values into (e), 
we have the value of V in known terms. 


Ex,—Find the angle of intersection of the two curves, 


RTE aS 


TANGENT CURVES. 39 
y? — dmx = 0, y? + (« —c)? —R? = 0, 
or of the two curves, 
y as, (y— 8) — mex = 0, 


PROPOSITION XXII. 

Determine the relation between the parameters of two 
curves tangent to each other. 

The constants that enter into the equation of a curve are called 
Parameters. Thus, if we have the equations, 

y—4mr=od, 
(F) yt+ae—Rh=o, 

(y—s) + (e — a)’ — Be ate0, 
m is a parameter in the first, R is a parameter in the second, and a, 
g, R are parameters in the third. 

We may designate in general that the equation of a curve con- 
tains any particular parameter, as R, by writing the equation thus, 
(115) 9(t, y, R) = 0, 
which signifies that the equation of the curve contains the variable 
co-ordinates, and the particular parameter R. Beside the particular 
parameter expressed, the equation may contain others not appearing 
in the general form of the equation. Thus (115) may represent 
either the second or third of equations (F). 

This premised, we proceed to the proposition. 

Let HL be one curve, and 
(116) (2, y, m) = 0 
its equation, in which m is a parameter. 


FIG. 18. 


Let AC be another curve, and Y¥ 
(117) F (x, y, n) = 9, 
its equation, in which mis a parameter. 
Suppose these curves tangent at the 
point P, <A line SR, tangent to AC at 
P, will also be tangent to HL at the R 


same point. 
If p and p' represent the differential coefficients of (116), and 


40 DIFFERENTIAL CALCULUS, 


(117), for the common point P, then since (by Proposition II.) p and 

p' each represent the tangent of the same angle R, we have 

(118) Pp =P’ 

in which it is to be recollected that p and p’ are functions of a and y. 
As p and p’ are functions of the co-ordinates of the point of tan- 

cency these co-ordinates x and y may be eliminated between (116), 

(117), and (118). The resulting equation solved for m may be re- 

presented by 

(119) m = f.n, 

which is the required relation between the parameters m and n. 
Ex.—Let the curves be the two circles, 

(a) a ie ad | a 

(b) (x — a®)+ (y— BY = m, 

| Equating the values of the differential coefficients of (a) and (6) 

| we have, after clearing of fractions, 

I (c) Gy sat, 

! Eliminate x and y from (a), (5), and (c), and we have for the re- 

lation between m and n, 


UN (d) m= (a? ate B?) + n. 

Equation (d) shows, that with the same centre there are two radii 
with which the circle (b) may be described so as to be tangent to 
the circle (a), one giving internal, the other external contact. 

ii From (d) we have 
ha (€) (a* + 8°)’ 
m) 1 
which shows, (since (a* + s°)? is obviously the distance of the 
centres) that when two circles have external or internal contact, the 
distance of their centres equals the sum or difference of their radii,— 

a proposition in plane Geometry. 

Determine the relation between the parameters m and n in the | 


: | 


bole 


[= 


t 


=> ANS Wy 


fines, 


y= ar+m yy=ar+m 
yim pagel or ) nto? + cat = a co, 
when they are tangent to each other. 
Equation (118) may be regarded as the condition of the tangency 


4 


GEOMETRICAL LOCI, = 2 


of two curves. ‘This condition results from the obvious principle, 
that if two curves are tangent to each other, they have a common 
tangent line at the point of contact. 


PROPOSITION XXIII. 
A circle touches two given curves, determine the Locus 
of its centre. 


FIG, 19, 


Let HC be one of the curves, 
and 
(120) 9(2', y') =o, R 
its equation, 

Let SR be the other curve 
and 
(Ket ie (i a ) 0, AN 
its equation. Let D 
(122) (x — a)? + (y—p)? = RY, 
be the equation of the circle tangent to (120) and (121) at P and Q, 
and in which a and 8 are the co-ordinates of the centre O. At P the 
co-ordinates of (120), and (122) are common. For that point (122) 
may be written, 
(123) (x7 —a)? + (y — sy? = R’. 

At (Q) the co-ordinates are common to (121) and (122). Hence, 
for that point (122) may be written, 


(124) (x'' — a)’ + (y’ — py? = R* 
Let p = the differential coefficient of (120) for the point P, 
= do. (123) P, 
P = do. (121) Q, 
P= do, (124) Q. 
Since (120) and (123) are tangent at P, we have, as in (118), 
(125) ay Oe 
For a similar reason we have, 
(126) ET et 


In these equations p and p’ are functions of 2’ and y’ and P and 
P’ are functions of 2'’ and y”, 


42 DIFFERENTIAL CALCULUS. 


The co-ordinates of the point P are common to (120), (123) and 
(125). Eliminating these co-ordinates between these three equa- 
tions, the resulting equation will contain a, 8, and R, and may be re- 
presented by 
(127) o(a, 8B, R) = o. 

The co-ordinates of the point Q are common to (121), (124), and 
(126). Eliminating these co-ordinates between these three equations 
the resulting equation will contain a, 6, and R, and may be repre- 
sented by 
(128) F (a, Oh) 0. 

Eliminate the variable radius R between (127), and (128). The 
resulting equation will contain a and gp, and may be represented by 
(129) pis, 

This being an equation between the co-ordinates of the centre, is 
the curve required. 

Ex.—Let the two given curves, HC and SR, be a circle, and 
straight line. ‘Then (120) and (121) become 


(c) xv? + y® — m = 0, and y'' —nz'' —b=o, 
the differential coefficients from which are 
(a) Cul, atmne said Qo» ere tpl 
dz' y da" 
The differential coefficients of (123) and (124) are 
(e) ay a ae ini, = P > and dy ee a re ige 
dx' y —8 dx y -—'p 
and equations (125) and (126) become respectively, 
yt x'—a x'—a 
— = ) and a= — ° 
(f) = = 


Eliminate x’ and y' between (123) and the first of (c) and (f/f), 
and we have the particular form corresponding to (127). Eliminate 
x'' and y'’ between (124) and the second of (c) and (f), and we 
have the particular form corresponding to (128). Eliminate R be- 
tween these two results, and we have for the locus of the centre a 
parabola,—a result which may be verified bys simple geometrical 


deduction, ~——-- ai 54 iat 


} 
; 


79 


~~ 


GEOMETRICAL LOCI. 495 


PROPOSITION XXIV. 
From a given point a line is drawn making a given angle 
with the tangent of a curve, determine the locus of the in- 
tersection. 


FIG. 2 
Let PD be the given curve, Se 
and Y | _ ae 
(130) o(7,y) = 0, | a hi 
its equation, PT 
Let M be the given point, and S Lae J M ID x 
(c,d) its co-ordinates, POPE ere Fae a BY, 


Let PS be the tangent line, and MH the line drawn through M, 
making at H the given angle with the tangent PS, 

Let V = tangent of the given angle at H, and let (m,n) be the 
co-ordinates of H,—the point of intersection. 

The equation of the tangent line PS for the point H is, as in (93), 
(131) a—y = p(m—x). 

The equation of MH passing through M is 
(132) y—d = x(2x'—c), 

For the point H this becomes, 
(132a) n—d = x(m—c). 

In these equations ~ is the tangent of the angle which HM makes 
with the axis of z. 


For the tangent of the angle H we have, as in (114), 

(133) at cd 
1+pn 

in which p is a function of «x and y. 

Eliminate x,y, and x between the four equations (130), (131), 
(132a), and (133), 

The resulting equation will contain no other variables than mand 
n. Representing it by 
(134) 1(m,n) = 0, 
we have the locus required. 

Cor. 1st.—If the angle H be a right angle, instead of (133) we 
have 


44 DIFFERENTIAL CALCULUS. 


(135) 1 + pz = 0, 
and 2,y, and x are then to be eliminated between (130), (131), 
(132a), and (135). 

Cor. 2d.—If the tangent of the angle H, instead of being con- 
stant, be a given function of the co-ordinates of the point H, i. e. if 
V = 9(m,n), 

then instead of (133) we have 
p—x 
1 + px 
and z, y and ~ being eliminated between (130), (131), (132qa), and 
(136), we have the locus required. 

Ex. 1.—Let the curve be a circle, and the point from which lines 
perpendicular to the tangent are drawn, be on the perimeter, find the 
locus of the intersection. 


(136) o(m,n) = 


FIG, 21. 


re 
f 


Take the origin A at the point 
through which the perpendiculars are 
| drawn. 
| Here the curve (130) is 
Ni (a) y* = 2rr—z’. & 
The tangent line HP is, as in (131), 

a (b) n—y = p (m—2). 

The equation of the perpendicular 
AH, passing through the origin and the point H, whose co-ordinates 
are m and n, is for the point H, 


< 


— 


‘ (c) n= xm, | 
and H being a right angle, we have, as in (1385), | 
(d) 1 + pr=o. 


Eliminate z,y, and ~ from (a), (5), (c), and (d), by the ordinary 
rules of Algebra, and the resulting equation is 
(e) n’ r? = (m* + n°) — 2rm (m* + n°), 
| the locus of the point H. This locus might be immediately obtained 
+ from the figure, by joining C and P, then CP is parallel and equal 
to SH, 
If (e) be changed to a polar equation, A being the pole, AH the 
radius vector R, and AX the angular axis, we have 


GEOMETRICAL LOCI. 45 


136f) R = 7 cos. a + 7, 

a very simple result, and which shows that if a semicircle be 
lescribed on the radius AC, and any line AS be drawn through A, 
and produced till SH equals the radius AC, a line through H per- 
pendicular to AH is tangent to the circle. As the curve (136f) 
possesses many remarkable properties, we will refer to it hereafter 
y the name of, Tue Crrctorp. 


- 


) 


Ex. 2.—Find the locus when the curve is an ellipse or hyperbola, 
and the point through which the lines are drawn is the focus, or 
centre, or vertex, the angle H being right. 

Ex. 3.—Find the locus when the curve is the parabola, and the 
point through which the lines are drawn is the focus or vertex, and 
the angle of intersection H is either right, or oblique. 

The properties of these loci are more readily investigated by 
-hanging their rectangular to polar equations, as was done with (e). 


PROPOSITION XXV. 
Pairs of tangents to a curve intersect at a given angle, 
letermine the locus of their intersection. 


FIG. 22. 


2 P 
Let PQ be the given curve, ¥ 


and PH and QH a pair of 


tangents to it intersecting at H. na 
Let (z',y') be the co-ordi- 
Pe 


, one of the points | 


d (z’’,y"’) the | | a 
0-ordinates of Q, the other D 


naies of f 


' tangency, an 


point of tangency. 


Let (m,n) be the co-ordinates of H, and let the equation of PQ 
for the point P be, 
o(z yy) = 9, 
i the equation of PQ for the point Q, 

) o(x’’,y P= oP, 


~ 
+ 


Let p and p’ be the differential coefficients of (137), and (138) 


; 
+) 


— 


= 
— ee 


a — 


: 

; 
1 ae 
{ / 


46 DIFFERENTIAL CALCULUS. 


The equation of the tangent line PH for the point H is asin (93), 


(139) n—y' = p(m—<’), 
The equation of QH for the point H is 
(140) n—y' = p' (m—2z’). 


Put V = tangent of the given angle H, then as in (114), we 
have for the tangent of the angle made by the lines, (139), and 
(140), | 
(141) PL AM 
| Ligh FP 

Since p is a function of 2',y', and p' of a'’,y', we have the five 
equations (137), (138), (139), (140), and (141), to eliminate the 
four co-ordinates of the points of tangency. The resulting equation 
will contain no other variables than m and n, and may be repre- 
sented by 
(142) Y(m,n) = 0, 
which is the equation of the locus required. 

Note.—If the curve (137) be of the second degree, the two values 
of each co-ordinate given, by the solution of (137) and (139), make 
known the co-ordinates of both points of tangency, and (138) and 
(140) are superfluous. This is indeed true, whatever be the degree 
of equation (137), for if (137) be of the n degree, then as was re- 
marked at the close of Prop. XV., the points of tangency are all 
given by the solution of equations (137) and (139). 

Cor. 1st.—If the angle H of intersection be a right angle, we 
have, 

(143) 1 + pp = 9, 
which is to be used in this case instead of (141). 

Cor. 2d.—If the tangent of the angle H, instead of being con- 
stant, were a given function of the co-ordinates of the point H, i. e. 
if V = o(m,n), we have, 


j eee i 
a 
which is to be used in this case instead of (141). 
Ex. 1.—Let the curve PQ be a parabola, what is the Locus of 
the intersections of pairs of tangents ? 


(144) o(m, n) = 


GEOMETRICAL LOCI, 


An? — 8rm 


(a) he 


(2m + r)? 
is the equation of the Locus determined as above directed, which is 
a hyperbola. If the angle be a right angle then (a) becomes 
(d) 2m + r= 0, 
which is a straight line—to wit, the directrix. 
Ex. 2.—Let the curve be the ellipse or hyperbola, and the angle 
H aright angle, what is the Locus of H? 


Ans. A circle. 
Ex. 3d.—Let the curve PQ be a parabola, and let the tangent of 
the angle H vary as a given function of the co-ordinates of H, 
what is the Locus. 
Representing the given function of the co-ordinates of H by 
p(m, n), we have, V = o(m, n). 


If we put this value of V into (a), we have 
1 


2 (n* — 2rm)y 

2m +r : 
which is the locus required. By giving any definite value to 9(m, 7), 
we may construct the curve (d). 


(d) g(m,n) = 


PROPOSITION XXVI. 


Pairs of tangents to a curve make angles with the axis 
of abcissas, whose product, or sum, or difference is constant, 
determine the locus of their intersections. 

The figure and solution is the same as in last Proposition, except 
that, instead of (141) we have in case of the product 
(145) Pps Cj 
a constant, and in case of the sum, we have 
(146) ptp =e, 
and in case of the difference, we have 
(147) p—p =e. 

Eliminate the co-ordinates of the points of tangency by means 
of equations (137), (139), and (145), and we have the locus when 
the product of the tangents is given. For the locus, when the sum 


48 DIFFERENTIAL CALCULUS. 


of the tangents is given, we have (137), (189), and (146), to 
eliminate the co-ordinates of the points of tangency, and when the 
difference is given we have (137), (139), and (147). 

Note 1st.—This proposition might have been more simply stated 
in the general form, thus: 

Pairs of tangents toa curve make angles with the axis of ab- 
cissas, any function of whose tangents is constant, determine the 
locus of the intersections. 

Designating the given function of the tangents by o(p, p’), we 
have by the proposition, 

(148) o(ps p!) = &. 

This equation embraces the particular cases (145), (146), (147), 
and indeed, (141), as also every possible combination of the tan- 
gents p and p’. 

Note 2d.—If the given function of the tangents, instead of being 
constant, varied as a given function of either or both of the co- 
ordinates of the point of intersection, we would have, when the 
function of the tangents varied as as a function of the abscissa, 
(149) o(p, p) = ¥ m, 
and when it varies as a function of the ordinate. 

(150) o(Psp) = 
and when it varies as a function of both co-ordinates, 
(151) Q(p.p’) = (m,n). 

We would then have (137), (1 39), and one of the three last equa- 
tions to eliminate the co-ordinates of the points of tangency. 

Ex. 1.—Pairs of tangents to a parabola make angles with the 
axis of abscissas, the sum of whose tangents varies as the abscissa 
of the point of intersection, find the locus of their intersection. 


Here (149) becomes 


(<) pap cm, 
where c isany constant, and (137), and (139) become & - 
(b) y? — 4ra = 0, and yn — 2r (m + 2) = 0. 


Eliminate x and y from the three equations (a) and (b), and we 
have for the locus required, 
(c) mc = n, a parabola. 

Ex. 2.—Pairs of tangents toa parabola make angles with the 


GEOMETRICAL LOCI, 


49 


axis of abscissas, the difference of whose tangents is constant, find 
the locus of intersection. 
Ans. A hyperbola. 
Ex, 3.—Pairs of tangents to a parabola make angles with the 
axis of abscissas, the sum of whose tangents varies as the ordinate 
of the point of intersection. Find the locus of intersection. 
Here (150) becomes 


pHi pesaken, 
and the locus is found to be a hyperbola referred to its asymptotes. 


PROPOSITION XXVII. 


A circle passes through a given point, and touches a given 
curve, determine the locus of its centre. 


FIG. 23. 


Let HD be the curve which is +| 
touched by the circle at P. 

Let B be the given point through 
which the circle passes, and (c,d) 
the co-ordinates of B, ‘Then we | 
have ~ Ma 
(152) (x—a)? + (y—p) = R’, 
for the equation of the circle in any position in which a and 8 are 
the co-ordinates of the centre, C. Let 
(153) O(2,y) = 0, 
be the equation of HD. 

At the given point B, whose co-ordinates are (c,d), the circle (152) 
becomes 
(154) (c—a)y + (d—p) = R*. 

Let us put p and p’ for the differential coefficients of (152) and 
(153), respectively. At the point P we have, as at (118), 

(155) p=P 
where p and p’ are functions of a and y. 

At the point P, x and y are common to (152), (153) and (155). 
Eliminate these co-ordinates between these three equations. The 
resulting equation will contain a,3, and R. Represent it by 
(158% F (a.8, R) = o. 


nel here SSS SSS 


Piha A 
wi th) 


50 DIFFERENTIAL CALCULUS. 


Eliminate R between (154), and (156), the result is an equation 
of the form 
(157) ¢(a,8) = 2, 
which is the equation of the curve required. 

Ex.—Let the curve HD be the circle, 

(a) BAe apPicm ig’, 
and let the given point B be at M, on the axis of abscissas. 

Equating the differential coefficients of (152) and (a), we have 
(0) ay = Bz. 

Eliminate x,y, and R between the equations (152), (154), (a) and 
(6). The resulting equation is, since d in (154) is, in this case, zero, 
(@)  @ (1S) 8 - See) —C ET =o, 

r ae Ar* 
which is a circle when c is zero, an ellipse when c is less than r, a 
hyperbola when ¢ is greater than 7, and a point, to wit, the centre, 
when c = 7. 


PROPOSITION XXVIII. 


Determine the maximum or minimum ordinate in a plane 
curve. 


FIG. 24, 


Definitions. 
1.—A maximum ordi- 
nate is greater than the 
ordinates immediately on 
each side of it. Thus P’D’ 
being greater than either 
of the ordinates, PD, or P’’D"’, is a maximum ordinate. PD and 


PD" are supposed to be indefinitely near to P’D’. 
FIG. 25. 


2.—A minimum ordinate is one 
that is less than either of the ordi- 
nates immediately on each side. 

Thus P'D' being less than either 
PD or P"D",is a minimum ordi- 
nate, PD and PD" being indefi- 
nitely near to P’D’. 


t 
— 


MAXIMA AND MINIMA, ' 


To determine the maximum or minimum ordinate, draw PR tan- 
gent to the curve at P. Then by (57) 


(158) @Y _ tan, PRX, 
dx 
Let the equation of the curve HC be, 
(159) y = pr. 


Suppose P’ be the vertex of the maximum or minimum ordinate. 

The nearer the point of tangency P is to P’, the further the an- 
gular point R is from the origin, When P’ becomes the point of 
tangency, the tangent line PR is parallel to the axis of abscissas, 
the angular point R is infinitely distant, and the angle P’RX is zero, 
or 180°. But when an angle is zero, or 180°, its tangent is zero ; 
consequently when P’ is the point of tangency, (158) becomes 
(160) dy cmt 

dz 

The two equations (159), and (160), suffice to determine x and 
y, the co-ordinates of P’. 

Ex. 1.—Let the curve HC be, 


(a) y= ar— 2x’. 
Then 
(b) dy  g —22. 
ax 
By (160) this is zero, i. e. 
(c) a—2r= 0. 
Solve (a) and (c) for x and y, and we have, 
2 
(d e=,and ys =. 
(¢) a? y 5} 


These are the co-ordinates of the point whose ordinate is a maxi- 
mum, and equation (d) shows the value of this maximum ordinate, 
or the co-ordinates of the point on the curve, whose ordinate is a 
maximum. 

This Proposition is merely a particular case of Proposition VIII. 
[t occupies, however, a large space in most books on the Calculus, 
und is worthy of particular attention. 


Determine the maximum ordinate in the following curves. 


DIFFERENTIAL CALCULUS. 
G) Sigil juaypik y= V— ea, 
\ y = 22° — az’, y? = ax — 2’, 
The following obvious principle extends this proposition to an ex- 


tensive and interesting class of problems. 
The same value of x which renders y a maximum or minimum 


Ha in such an equation as (159) will render it a maximum or mini- 
val mum whether (159) be the equation of a curve or have any other 
i q y 

wil signification. 

el Hence, the principle of this proposition serves to determine the 
Baty maximum or minimum value of any expression involving a variable 
Tae quantity. Illustrative of this take the following problem : 

fi i) Ex. 2.—Divide a given straight line into two parts whose rectangle 


Bai) will be the greatest possible. 
| i FIG. 26. 


i Let AB = a, the given ; ' 
rn line, C the point of division. - c 7 
l AG ear iste CB = te a, 
ull By putting y for the rectangle of the parts of the line, we have, 
Sai (e) y = ar — 2; 
| ify the same as equation (a) above. Differentiate (e) and equating its 
i differential coefficient to zero, as in (c), we have, 


it (f) c= > 


1 dba which shows that the line is bisected. 
itl As another example, take the following problem : 
Ex. 3.—Though a given point in a given angle draw a line, so as 


fi) to form with the sides of the angle the least possible triangle. 
FIG. 27. 


Let A be the given angle. P 
nae the given point. 
Ae AD =a? given co-ordinates 
a yitus oD =b of P. 

Put DE = a, and by similar 
triangles, we have, 


(g) x 


a 
& 
-b 
Q 
> 
an 
| 
alo 
“ 
=m 
= 


MAXIMA AND MINIMA, 53 


Putting y for the area HAE, we have, by the rule for the area of 
a triangle, 
a+e2y Osin. A 
(i) Y= oat nell tami 
Z 2 
Differentiate (hk) and putting the differential coefficient equal to 
zero we have, 


9 heey »\2 i f 
(k) ot as i lit Pes 


2 . 
ae Pa 


Solve (k) for x, and we have, 2 = a, which shows that the area 
of the triangle AHE, is the least possible when AD = DE. 
Comparing (h) and (k) we observe that the constant factor, 
b sin, A 


? 


») 
remains in differentiating, and divides away when the differential 
cofficient is equated to zero, This factor might, therefore, be omit- 
ted before differentiation, without affecting the result. 

Again, since any power or root of a quantity is obviously a maxi- 
mum when the quantity itself is a maximum, the power or root may 
be omitted before differentiating. Thus, (ax — 2)", is obviously a 
maximum when az — 2 is a maximum, These considerations 
might be expressed in the foilowing rules: 

ae 

A constant factor of a quantity to be made a maximum or mini- 

mum, may be omitted before differentiating. 


B. 

If the quantity to be made a maximum or minimum, be all under 
a given power or root, the power or root may be omitted before dif- 
ferentiating. 

C. 

Since the Logarithm of a quantity increases or decreases as the 
quantity itself increases or decreases, an expression to be made a 
maximum or minimum, may be put into Logarithms before differ- 
entiating. 

D. 
When the differential coefficient is zero, the differential is also 


= 


f 


04 DIFFERENTIAL CALCULUS. 


zero. Hence when a quantity is a maximum, or minimum, its dif- 
ferential is zero. 


E.. 

In solving problems in Maxima, or Minima, the first process is 
to obtain an algebraic expression for the quantity to be made a max- 
imum or minimum. This algebraic expression will be a function of 
some variable. Differentiate this expression and put its differential 
equal to zero. ‘This furnishes an equation from which the value of 
the variable of which the expression is a function, may be obtained. 

The determination of the expression for the quantity to be made 
a maximum, or minimum, is frequently the most difficult part of the 
process. ‘This is, however, the work of Algebra and Geometry. 
The Differential Calculus lends its aid only after this expression is 
obtained. 

Illustrative of the foregoing rules and remarks, take the following 
example. 

Ex. 4.—Cut the greatest ellipse from a given cone. 

FIG, 28, 

Let CAB be the given 

cone. Put 
c = cosine of C, 


and 

a=AQC, b= AB, x—CE. 
AE is the trace on the 

plane of the paper, of the 

plane that cuts the cone. 

AE isthe major diameter TF 


of the ellipse, and AER the {= é ev 


half of the ellipse. ioe eee 

Cut the cone by a plane passing through O, the centre of the 
ellipse, and parallel to the base. The section is a circle, the half 
of which is HRG. This circle passes through the extremities of the 
minor diameter of the ellipse. Hence OR perpendicular to the plane 
of the paper, at O, passes through the point R, the intersection of 
the ellipse and circle, and OR is the semi-minor diameter. 


vi 


MAXIMA AND MINIMA, 5s 
By Trigonometry, we have, from the triangle CAE, 
1 
(7) AE = (@ + 2? — 2 caz)’, 
and from the triangle CFE, we have, 
1 i 
(m) EF = (CF*4+ CE*?— 2CF.CE cos. C)? = x (2— 2c)’. 
Since AO is the half of AE, we have, from the similar triangles 


AEF and AOH, HO = EF ~ 2, or 


l 
) 


r 
n HO = — (2 — 2¢ 
(n) 5 (2 = 2¢). 
Also, by similar triangles, EOG and EAB, we have, 


Since RO is an ordinate of the semi-circle HRG, we have, by 
Geometry, and the values (n), and (0), 


(p) OE 


9) 
~ 


(2 men Se)", 


Equations (/) and ( p) make known the diameters of the ellipse. 
The rectangle of the diameters multiplied by a constant 7 — 4. is 
the area of the ellipse. Hence we have, 


i 


(7) “. areaofellipse = — b} (2 —— 8a) (Wx + 2° — 2caz’)*. 


So far the process is one of Algebra and Geometry. To apply 
the Calculus, omit the constant factor in (q), and the radical, and we 
have, 

(r) ax + x — 2caz’, 
to make a maximum, 

Put the differential of (7) equal to zero, and solve the equation for 
x. We then have, 

(s) = rr ac=s Per (4c° “oan BY 
3 3 

Equation (s) determines the position of the vertex E of the 

ellipse required. If , 


1 
‘Nee 3\2 40 eee 
4c? = 3, then C= (—) , and the angle == 30 


F Shi! 
I 


hil ii 
MeL 


56 DIFFERENTIAL CALCULUS, 


If in (s) we have 4c? less than 3, or the vertical ang.e greater 
than 30°, x is imaginary, or such a cone does not admit of a maxi- 
mum ellipse. 

We will revert to this example in a subsequent Proposition. 

We will add a few more examples of a general nature, under this 
Proposition. | 

Ex. 5,—Determine the point on a curve through which if a tan- 
gent line be drawn, the triangle formed by the tangent and co- 
ordinate axes will be a minimum. 


Let PD be the curve, and Ayo P 
és 


PC the tangent line required. 
Then AHC is the triangle whose 
area is a minimum, and P is the 
point to be determined. 

Let (a, y) be the co-ordinates 
of P, and represent the equation of PD by 
(2m) Yy = ox. 

Then as in Proposition XX, (110), (111), we have for AC and 
AH, 
(27) AC = «— a and AH = 9a — a2. 

Then putting A for twice the area of AHC, we have, by the rule 
for the area of a triangle. 


(2p) Aves (« — =) (p% — xyz), 


Here the area is a function of 2, the abscissa of P. Differentiate 
(2p) and this differential put equal to zero (by Rule D, above) fur- 
nishes one equation which, with (2m), the equation of the curve, 
makes known the co-ordinates x, y, of P. 

In applying this process to any given curve it would generally 
be more simple to get, first the equation of the tangent line for the 
particular curve, and then get the intercepts AH, AC, and form the 
area AHC, which may always be expressed in terms of one variable 
by means of the equation of the curve, whether that equation be ex- 
plicit or implicit. 


MAXIMA AND MINIMA. oF 


Ex.—Draw a tangent line, cutting off with the axes, the least 
triangle, when the curve is the circle with the origin at the centre. 
Here (2m) becomes 


(2q) a+ y—h’ =o, 
and for the tangent line to the circle, we have, 
(27°) vatyy—R=a, 


From (27) find the length of the intercepts AC and AH, and we 
have for double the area of the triangle, 
> 
A — Rt 
xy 
from which one of the variables 2 or y may be eliminated by equa- 
tion (2q), and the differential being put equal to zero, furnishes one 
equation, which with (2q), makes known a and y, the co-ordinates 
of the point required. In this example # and y are found to be 
equal. 


Ex.—Draw such a tangent line to the curves, 


ey+Pwv—av’h=o, vy—VPxevridl=o. 
Ex, 6.—In a given curve inscribe the greatest rectangle, 
FIG. 30. 


a 


Let (x, y) be the co-ordinates 
of P, and, 
(t) y = 92, 
be the given curve AH. 
et AS=a, a known length, 
ES = a—xz. 


Then the area of the rec- 


tangle PS, is obviously, ee E S 
(u) PS = (@— 2) y, 

or putting for y its value in (¢t), we have, 

(2u) PS = (a — @) oa. 


Differentiate (2u), and its differential put equal to zero, furnishes 
an equation which, with (¢), makes known the point P, and deter- 
mines the rectangle. 

Determine such a rectangle in the curves, 


| 


1 
y= ay, y= (mz), y= (R? —- a}. 
ig 


Pe teeny 


sSoraiiee mcd a ane 


58 DIFFERENTIAL CALCULUS. 


Ex. 7.—Determine the greatest cylinder that can be inscribed in 
a given surface of revolution. 

If the curve AH, and the inscribed rectangle PS, revolve round 
AX, AH will generate a surface of revolution, and PS will generate 
the inscribed cylinder. The radius of the base of the cylinder is 
obviously PE, Let (a, y) be the co-ordinates of P. Then x y? ex- 
presses the base of the cylinder, and the solidity of the cylinder is, 
(v) cylinder = x ¥? (a — z) = x (9 x)? (€ — 2). 

Equate the differential of (v) to zero. This equation, with (¢), makes 
known the point P, which determines the cylinder. Determine such 
a cylinder when the surface of revolution is generated by any of the 
curves, 

Pt+y—R=o0 y=ar, y= pr, ?y +02 —v’v=o. 

Before leaving -this proposition, we will explain a process which 
is often found convenient. 

Resume the simple example :—Divide a given straight line so 
that the rectangle of the parts will be a maximum. Putting a for 
the length of the line, x and z for its parts, and y for the rectangle, 
we have, 

(w) I me 1 AR 
(2a) fF hs 

If we eliminate z from (2a), by means of (w), we have equation (e), 
as before. But, instead of first eliminating z from (2a), differentiate 
(2a,) and since (2a) is a maximum, we have, 

(2b) adz + zdx = o. 

Differentiate also (w), and we have, 

(2c) » Or -- dz = 0, 

By means of (w) and (2c) eliminate z and dz from (2b). The 
resulting equation will contain dz in every term, which being divided 


away and the equation solved for x, we have, 
a 
L ae 
2 
This process may always be followed when the expression to be 
made a maximum or minimum contains two variables, as in (2a). 


and these variables are related in another equation, as in (w). 


SECOND DIFFERENTIALS. 59 


The nature of the problem, in most cases, indicates whether it is 
a maximum or minimum, which is given by the foregoing processes, 
An analytical test will be given in a subsequent proposition, by 
which the maxima can be distinguished from the minima values, 

When the ordinate admits of neither maxima nor minima 
values, the algebraic expression itself generally makes known the 
fact, or it can be readily inferred from the equation obtained by 
equating the differential coefficient to zero. Thus, if the example 
were, 

Find the maximum ordinate in a parabola, 

the equation of the parabola, y* = 4m, shows that y is greatest 
when 2 is greatest, and least when 2 is least ; consequently there is 
no ordinate which is greater or less than the ordinate immediately on 
each side of it. If we differentiate this equation, its differential co- 
efficient put equal to zero, is 


2m m 
ete Fe = 0, OF ee OC. 


y x 

The first requires y to be infinite, the second requires 2 to be in- 

finite. Consequently, according to the definition, there is no maxi- 
mum or minimum ordinate in the parabola. 


PROPOSITION XXIX. 
Given the equation of a curve, determine its second dif- 


ferential. 
FIG. 31. 
Let BR be the given curve, Y 
and 


(161) y= az’ + Bb, 
its equation, in which let 2,y be 
the co-ordinates of P. 


We have seen, (equation (1) to A 
(9), Proposition I.,) that the differ- EMN 
ential of (161) is 
(a) aye sax4) dx, 


where dy may be represented by QO, and da by EM. Make 


60 DIFFERENTIAL CALCULUS. 


EM = MN = da. Put y' = QM, then AM, the abscissa of Q, is 
x+ dx. 

We observe in (a), that the differential of the ordinate of BR, for 
the point P, equals the constant 3a into the square of the abscissa 
of P, into da. 

In like manner, the differential of BR, for the point Q, is 
(5) dy’ = 3a (x + da) dz, 
where dy’ may be represented by RC, and dx by MN. 

Expand (6), and subtract (a) from it, and we have, 

(c) dy' — dy = 6az.dx*? + 8a.dz°, 

The first side of (c) is the difference of the increments of two or- 
dinates, indefinitely near to each other, and is what is understood by 
a second differential. 


Let us represent dy’ — dy by dy, which is read, The second 
differential of y, and then dividing (c) by dx’, we have 
d’y 


(d) _ 4% = 6axr + 8a.dz. 
dx? 


Now in the limit, or when dz is indefinitely small, the second side 
of (d) reduces to the term 6ax, and (d) becomes 


(e) ay == Oar 
dx? 
If this be cleared of fractions, we have, 
(f) We) ==) OO2,0ane 


Equation (f) is the second differential of (161). It is obvious 
that (f) is the differential of (a), on the hypothesis that dz isa 
constant. But in these processes, a is the independent variable. 
Hence to obtain the second differential, differentiate the first dif- 
ferential, making the differential of the independent variable 
constant. ‘ 

Ex. ics Aah) dy = naz" dz, and 

dy = n(n —1) az” da’. 

In like manner, we would get the third differential by differentia- 
ting the second differential, making dx constant, The forms d°y, 
d'y, d”y, are used to denote the third differential, the fourth differen- 
tial, é&c. of y. 


SECOND DIFFERENTIALS. 61 


Equation (e) is called the second differential coefficient. 
Differentiate (e), and divide by dx, and we have, 


3 
(g) ay = 6a, 
dx?’ 


the third differential coefficient. 

Differentiating (g), and dividing by dx, we have, 
(h) ——~ on 

Hence the fourth differential coefficient of (161) is zero. 

Cor.—It is obvious that between (161), (a), and (f), we could 
eliminate the two constants a, and 3, that enter into (161). The 
result of this elimination would be, a differential equation of the 
curve, involving no constant, It is also obvious that if (161) con- 
tained three constants, we would have to employ the first, second, 
and third differentials to eliminate them, &c. 


Eliminate a between equations (161), and (a), and we have, 


ay x 
k y oe SPER ore: + b, 
(*) dx 3 


which is also the first differential of (161). 

Kliminate @ and 6 between (161), (a) and (f), and we have, 
yin dy 
dx 2 dz 

which is, like (f), the second differential of (161), 


(2) 


PROPOSITION XXX. 


If a curve be concave towards the axis of abscissas, its 
second differential is negative. If it be convex towards the 
axis of abscissas, its second differential is positive. 

First.—When the curve PL is concave towards the axis of ab- 
scissas, let P be any point on it. 

be 


DIFFERENTIAL CALCULUS. 


FIG. 32, 


Let PO = QC = de 

Join the points PandQ,and y 
produce PQ to meet SC in H. It 
it obvious that PQ must rise 
above the curve at Q. H is 
therefore above 8S, and SC is less 
than HC. ButsincePO=QC, A 
we have HC = QO. Also, dy' — dy, or its equal, d’y, as used in 
the previous Proposition, is HS, that is, 


(161la) d?y = SC — QO = SC — HC = — Us. 
Hence when the curve is concave, the second differential is 
negative. 


Second.—When the curve PL is convex towards the axis of ab- 

scissas, let P be any point on it. 
FIG. 33. 

Make PO = QC = dx. vy 

Join P and Q, and produce 
PQ to meet SC in H. 

At Q, PQ descends below the 
curve, Reasoning as before, 
we have aan 
(1615) dy = SC — QO = + SH. 

Hence if the curve be convex towards the axis of abscissas, the 
second differential is positive. 

Ex.—lIs the curve, y = ax + 2’, concave or convex towards the 
axis of x? 


Ans.—Convex. 

Cor. 1st.—As the differential coefficient has the same sign as the 
differential, the second differential coefficient is negative when the 
curve is concave, and positive when the curve is convex towards the 
axis of 2. 

Cor. 2d.—This Proposition enables us to determine when a curve 
given by its equation is concave or convex towards the axis of <, 
and also gives us the means of determining the point where it 
changes from convex to concave. 


SECOND DIFFERENTIALS. 63 


FIG. 34. 


Thus let the equation of the 
curve SQ be, 


(c) y = az’ — 7° + 5B, 
The second differential co- 
efficient of this is D E xX 
d. 4 
(d) std = 2a — 62, 
dx? 


Equation (d) shows the second differential coefficient to be posi- 
tive, when 2e@ is greater than 62, and negative when 2a is less than 
6x. Consequently the curve SQ is convex for all values of « less 
than a + 3, and concave for all values of x greater than a ~— 3. 


curve changes from convex to concave. 

The point R, where this change occurs, is called ¢ singular point, 
and is evidently determined by putting the second differential co- 
efficient of the curve equal to zero, for when a equals a ~ 3, the 
second side of equation (d) is zero, 

And in general, if the equation of SQ be 
(161k) Rona” Oe 
this equation, in conjunction with the equation, 


(161k) —_ = 0, 


must determine the co-ordinates 2 and y of the singular point, if the 
‘urve have such a point. If the value of the second differential co- 
efficient, from (161k), does not contain either of the co-ordinates, it 
is obvious that (161k) does not exist. Thus, to illustrate by a par- 
ticular case, if the equation of the curve be, 


(e) nd | a ee 
then the second differential coefficient is 
£ d’y aie 7 | 
(J ) > ae te 
iw 
If this second differential coefficient be put equal to zero, we 
have 2 = 0, which is absurd. Hence the curve (e) has no singular 


point, 
Ex.—Determine the singular points of the curves, 
y = ar’ — z, == ax —~ 24, YS 1 — Be 


) 


Hig by 


be 
iat 
hil 


jh 


DIFFERENTIAL CALCULUS. 


PROPOSITION XXXI. 


When a curve has a maximum ordinate, the second dif- 
ferential coefficient is negative, and when it has a minimum 
ordinate, the second differential coefficient is positive. 

It is obvious that when a curve is concave towards the axis ol 
abscissas, it may have a maximum ordinate at some point L, (see 
Fig. 32, last Proposition,) but cannot have a minimum ordinate. In 
Proposition XXX., it was shown that when the curve is concave to- 
wards the axis of abscissas, the second differential coefficient, at any 
point of the curve, is negative. Consequently at the point whose 
ordinate is a maximum, the second differential coefficient is negative. 

It is also obvious, that when a curve is convex towards the axis 
of abscissas, it may have a minimum ordinate at some point L; (see 
fig. 33, last Prop.) but cannot have a maximum ordinate. Conse- 
quently, by what was shown in Proposition XXX., at the point 
whose ordinate is a minimum, the second differential, and second 
differential coefficient are negative. 

These principles afford an obvious test to determine the maxima 
and minima values of algebraic expressions, and enable us to dis- 
tinguish the one from the other. 

(a) Ex. 1.—Does the curve y = ax — 2’, admit a maximum 
or minimum ordinate ? 

Here the first differential coefficient 1s, 


(5) 


which, put equal to zero, gives, 
a 
ih 
The second differential coefficient is, 
2 
() oY = —2, 
dx 
which being negative for all values of x, shows that the curve ad- 
mits of a maximum and not of a minimum ordinate. 
If the second differential coefficient (c) contains «, the values of 


cc 


SECOND DIFFERENTIALS. 65 


x derived from the first differential coefficient put equal to zero must 
be substituted into (c). If the first differential coefficient put equal 
to zero, furnishes an equation above the first degree, the values of 
the variable x derived from it must be severally substituted into (c). 
Some of these values may render (c) positive, and others of them 
negative. In such a case the curve admits of both maxima and 
minima ordinates. 
(d) Ex. 2. y= axr—2+ 8B, 

The first differential coefficient of (d) put equal to zero, gives the 
values, 


(e) z= + (5): and 


(f) ce (<)- 


Also, the second differential coefficient of (d) is, 
. d’y 
(g) dak 


If in (g) we substitute the value of 2 in (e), the second differ- 
ential coefficient is negative. If in (g) we substitute the value of 
x in (f)), the second differential coefficient is positive. 

Hence, (d) has both a maximum and a minimum ordinate. The 
maximum ordinate is given by the value of 2 in (e), and the 
minimum ordinate by the value of x in (f). 


{~ 


bo 


— — 62, 


& FIG. 35. 
The form of a curve that has 


both a maximum and minimum 
ordinate, may be seen in fig. 35, 
in which AS would be a mini- 
mum, and EQ a maximum or- ' 
dinate, D EK x 
Such a curve must be convex where the ordinate is a minimum, 
and concave where it is a maximum; consequently, every curve 
that has both a maximum and minimum ordinate, must have a sin- 
gular point between the maximum and minimum ordinate points. 
From the same figure we may infer generally, that when there 
are two values of x given by the equation made by putting the 


Y 


68 DIFFERENTIAL CALCULUS. 


S 


differential coefficient equal to zero, one of them gives a minimum, 
and the other a maximum ordinate. For, itis obvious in the figure, 
that there cannot be two maximum ordinates without an intervening 
minimum one, nor the reverse. Since any algebraic or transcen- 
dental expression may be regarded as some power of the ordinate 
of a curve, the principle of this proposition serves to distinguish 
the maxima and minima values of all algebraic or transcendental 
expressions. 

As an illustration, recur to Example 4, Proposition XXVIII, 
which proposed,—7'o cut the greatest ellipse froma given cone. 
Putting y equal to the expression (7), in that proposition we have for 
the second differential coefficient, 


(h) ok OF: — Aca, 


And the two values of (s) in that proposition may be written 
separately. 


cs 

9) 2 

(k) Been += (4c? — 3). 
; 5 ln, oh 9 
(¢) SRE OF agen! LEC Tet) 


If for x in (h) we substitute its value in (A), the second differ- 
ential coefficient is positive, and if for a in (h) we substitute its 
value in (J), the second differential coefficient is negative. Hence, 
the value of « in (k) gives a minimum ellipse, and its value in (0), 
a maximum ellipse. 

If the equation of the curve be, 


it might occur that the values of x deduced from 
dy d’y 

' eS and: 5 

(7) aa Oand 0 


would be the same. 

When this is the case the tangent line to the curve at the singular 
point is parallel to the axis of 2. For the first of (n) existsatthe 
point whose tangent is parallel to the axis of x, and the second of 
(n) exists at the singular point of the curve (m). These points co- 
incide when (m) and the first of (n) give the same values of a and 


~ 


CIRCULAR FUNCTIONS, 67 


y that are given by (m) and the second of (nr). If the two equations 
(n) exist, but not together, the singular point does not coincide with 
the point whose ordinate is a maximum. 

Eix.—Determine the maximum ordinate and singular point of the 
curve whose equation Is, 
(p) y = ax’ — x. 

The first and second differential coefficients of this put equal to 


zero we have the equations, 


di an se 2a 
(q) uy = 2ax — 32? = 0; “en: == O,Or ¢ = —— 
dx 3 
| ad”) ee 4 a 
7) J — 24 — 62 = 0 olf a pcieaed 
dx? 3 


Equation (q) gives two values of 2, and equation (7) a value of 
x different fromm both values in (q). . Hence, the singular point in 
(p) does not coincide with the point whose ordinate is a maximum 
Or a minimum. 

The value « = o gives the minimum ordinate AS and the value 
x = @ — 3, gives the singular point R; and the valuea = 2a 3 
gives the maximum ordinate QE. 


PROPOSITION XXXII. 


Determine the differential of a circular function. 

We defined a circular function to be one involving the variable in 
an arc or trigonometrical line, as sin.z, cos.2, &c. We will first 
determine the differential of the sine of an arc. 


FIG. 36. 


Let the are AB = x, BC = y, 
and OA = 1 the radius, 


(162) me = Silt 
Suppose the arc AB receive the 

increment BS = h, Put y' = SL, 

and we have, 

(163) y = sin.(x + h), 
Expand (163) by the usual tri- 

gonometrical development, sub- | 

tract (162) from it, divide by h O@————-——_+_4+—A 


and we have, 


68 DIFFERENTIAL CALCULUS. 


y —y _ sinh. ee ee cosh sin.x 
h h h h 
When an arc is indefinitely small it may be regarded as equal to 
or coinciding with its sine, and its cosine may be regarded as equal 
to or coinciding with the radius, therefore, when the increment h is 
indefinitely small, we have, 
sinch = h,Y cos. 1" radius, 4 —"y"= dy he az: 
These conditions reduce (164) to 


(165) Ppa COS.2, 
dx 


(164) 


or multiplying by da, 
(166) « dy = cos.z.dz. 
Equation (166) is the differential of (162). Comparing (162) 
and (166), we have, 
RULE X. 
The differential of the sine of an arc is the cosine of the arc into 
the differential of the arc. 
Ex.— y = sin.mz “. dy = cos.ma.mda, 
yp = Cy == n'sIn."—'2.008 Tax, 
y = sin.(m x + nx"), - Con 
Next let us determine the differential of the cosine of an arc. 
By trigonometry, we have, 


(a) sin’a + cos2a = 1. 
Differentiate this by previous rules, and we have, 
(d) 2 sin.x cos.xdx + 2 cos.xd.cos.x = 0, 


where d.cos.a designates that cos.a is to be differentiated. 
By transposition and division, we have from (0). 

(c) d.cos.x = — sin.xdx, 
Equation (c) furnishes 


RULE XI. 
The differential of the cosine of an arc is minus the sine of the 
are into the differential of the arc. 
Ex.— y = cos.mz . dy = — sin.ma.mdx, 
fe COS," “. dy = —n cos.” 2,sin.edz, 
y = cos.(mx + c). / 


CIRCULAR FUNCTIONS. ot 


To differentiate the versed sine of an arc, we observe by figure 


36, that 


(d) versin.2 = 1 — cos.2. 
Differentiate this, and we have, 
(e) d. versin.w = sin.rdz, 


which furnishes the proper rule. 
To determine the differential of the tangent of an arc, we know 


that, 
“ sin.x 
( f ) tan.2 = . 
: cos.a 
Differentiate ( f) as a fraction and reduce by (a), and we have, 
dx 9 
(g) d. tan.x = ———— = secan,’«. dz, 
a cos.°x 


which furnishes the rule. 
Exs. y= tan. (2? + a) . dy = secan.*(2* + a). 2xdz. 
y = sin. tan.2, 
Such an example is differentiated as the product of two factors, 
sin.x being one factor, and tan.x the other. 
We have hitherto supposed the trigonometrical lines to be func- 
tions of the arc. Let us reverse this process, and consider the arc 


as a function of the trigonometrical line. 
FIG. 37. 


Let AB = 1, the radius. Y 


Let the arc BC = y, and the sine line 


me 
TD : é . C 
CE = 2, the cosine line AE = z, and 
the tangent line BP = u. Then 
(A) 2 = siN.y, 4 = cos.y, u=tan.y. 
Eb 2 


In order to express the arc y, in terms 
of the trigonometrical lines, the equations 
(h) may be written, sr 
(k) acm, SiN y = cos.—'z, and y= tale 
where sin.—'x is a mere arbitrary notation employed to designate the 
arc of which = is the sine line. [n like manner, cos.—’z designates 
the arc of which z is the cosine line, &c. ‘To differentiate (x), take 
the direct forms, (h). From the first of (h), we have, 

Q x 


70 DIFFERENTIAL CALCULUS. 


1 
(Z) Gy = i But cos.y = AB *=)(1—2*). 
COS.¥ 
Hence, 
(m) Feet RES: 2080 
(1 — 2)" 


Equation (m) furnishes the rule for differentiating the first of (i), 


VIZ; 


RULE XII. 

The differential of an arc, in terms of the sine line, is the differ- 
ential of the sine line, divided by the square root of radius squared, 
minus the sine line squared. 

Proceed in the same way with the second of (h), and we have, 

dz 


(7) dy = —- us 
(1—2’)* 


which furnishes the rule for differentiating the second of (kk), viz: 


RULE Xiil. 

The differential of an arc in terms of the cosine line, is minus 
the differential of the cosine line divided by the square root of 
radius squared, minus the cosine line squared. 

Proceed in the same way with the third of (h), and we get, 

du 
1 + uw 
which furnishes the rule for differentiating the third of (k), viz: 


(0) a9 


RULE XIV. 


The differential of an arc in terms of the tangent line, is the dif- 
ferential of the tangent line divided by the radius squared, plus the 
tangent line squared. 


If v = the versed sine EB, then we have, 
(p) » = versin.y, and y == versin. 6. 
Differentiate the first of (p), and we have, 
dv dv 


(9) By es hr ee 


CIRCULAR FUNCTIONS. 71 


which {furnishes the rule for differentiating an arc in terms of the 
versed sine. 
Exs,—Differentiate the equations, 


y = versin.—'. av, y = versin.—'v’, 
y = tan.-'\(@% + a), y = tan. (x + a)’, 
y = cos.'(% + @), y = sin.’ (x + a). 


[Illustrative of the preceding rules, we subjoin the following ex- 
amples, which are solved by the principles of Proposition XXVIIL,, 
and in which circular functions are employed. 

oy. aa 

Divide a right angle so that the rectangle of the sines of the parts 
will be a maximum. 

Let x equal one part of the angle, then 90° — a, is the other, 
and we have, 

(7) Rectangle = sin.a. sin.(90° — x) = sin.@ cos.x. 

Differentiate (r), as the product of two factors, by Rules X. and 
XI., and putting the differential zero, we have, 

(s) cos.’a.dx — sin’Zx.dx = o, 

From this we have, 

COs. == S1N.7, 
which shows that the angle is bisected. 
Ex. B, 

In a given semicircle, inscribe the greatest isosceles triangle, one 
of whose sides coincides with the diameter, and whose vertex is on 
the perimeter of the circle. 

FIG. 38. 

Let the radius of the circle be, 

OA = 1, and let CAD be the Cc M 
required triangle. 

Let the arc ASC = 2z. Let 
AC and AD be the sides of the 


isosceles triangle. Put Bp O A 
(¢) y= AC=AD, .-.- y =2sin.z,and CAB = 90°—z. 
(u) “. gin.CAB = cos.z. 


Put z for the area of the triangle, CAD, and we have, by the rule 
for the area of a triangle, 


~] 
dN 


DIFFERENTIAL CALCULUS. 


Bi -WCOS 2? Wie aru ’ 
(?) a= yp. Mite ss. 
2 
Substitute into (v), the value of y in (¢), and we have, 
(w) % = 2 sin.*2, cos.e = 2fcos.x — cos.°x 


Differentiate (w) by Rule XL., a the differential han pier 
to zero, and solved, we have, 


i 
(2) DOS of == (=): or SI.2 == 1D, 


The first gives a maximum, the second a minimum triangle, as 

may be determined by Proposition XXXII. 
Exc, 

In a given semicircle, draw a chord AC, (Fig. 38), such that if a 
semicircle CMA be described on the chord, the lune CMAS will be 
the greatest possible. 

Taking the same data as in Ex. B, and putting x for the area of 
the lune, we readily obtain from Plane Geometry, and Trigonometry, 
the expression for the area, viz: 


(2a) = 1 nsin2@ + sin.x cos.2 — 2. 

The differential of (2a), being put equal to zero, we via 
(25) 7 SiN. Cos.% — 2 sin.’x = 0, 

This is satisfied by making, 
(2c) sin.2 = 0, OF x cos.c — 2sin.z = o, 


The first gives a minimum, the second a maximum lune, as may 
be determined by Proposition XXXI. 
In the deduction of the previous rules, we supposed the radius of 


the circle to be a unit. 


FIG. 39, 


H 


Suppose, however, we 
have to employ an are HE, 
whose radius is AH = R., 

Take AG, a radius equal 
to unity, and describe an arc 
GC, Draw the sine lines 
HD and GB of these arcs. 

Put =2z,and HD=u. 


;™ 


, 


POLAR SUBTANGENT. 703 


By similar triangles, AGB and AHD, we have, 
(2d) Via y Rb, 


from which we get, 
(2e) ote Tee 7 


Also, since similar arcs are as their radii, we have, from the sim- 
ilar arcs GC and HE, 
(2f ) I3.GC ssahoeei, 

By the notation for an arc in terms of its sine line, we have, from 


the fi 


Ce 
>° 


: ; u 
(22) GC = sin. zr = sin. R: 
is 


Substitute this value of GC into (2f), and we have, 
(167) HE = R sin —. 
R 

Equation (167) expresses that the arc, whose radius is R, is b 
times the similar arc whose radius is unity. 

If z be the tangent of the arc HH, we have, in like manner, 


(168) 4 ° HE = R tan-' 7, 
R 


Proceeding in the same manner, we would obtain similar values 
when the arc is a function of any other trigonometrical line. 


PROPOSITION XXXIII. 


Determine the polar subtangent of a curve. 

Definition —If a line be drawn through the pole perpendicular to 
the radius vector to the point of tangency, the part of this line be- 
tween the pole and tangent is the polar subtangent. 

The method of changing the equation of a curve from rectangular 
to polar co-ordinates, is explained in Analytical Geometry. Many 
curves, as the spinals, &c., are more readily discussed by referring 
them to polar co-ordinates. 


74 DIFFERENTIAL CALCULUS. 


FIG. 40. 


Let O be the pole, OE the an- 
gular axis, OP the radius vector, P 
the point of tangency, PR the curve 
referred to polar co-ordinates, PD the 
tangent, and OD perpendicular to 
OP, then is OD the polar subtangent. 

To find the length of OD, in terms 
of constants, and the co-ordinates of 
P, take the radius OF = 1, and des- 
cribe the arc EL. 

Let the point Q be indefinitely near 
to P, draw OQ, and with centre O, 
and radius OP, describe the arc PS, 
then we have, OS = OP. D 

In the limit, or when PQ and PS are indefinitely small, they may 
be taken as straight lines, and the angle S may be taken a right 
angle. Since EH is the measure of the angle POE, put 
(A). EH = Ws OP = R. Then HG = da, SQ = dR. 

Since similar arcs are as their radii, the ares HG and PS give, 
(169) 1:do::; R: PS = Rdao. 

The tangent PD may be regarded as coinciding with the curve 
along the indefinitely small arc PQ, and the triangles QSP and 
POD are similar, from which we have, 

(170) OS 2 SP 22)PO. OL 

By means of (169), and (A), this becomes, 
(171) dR’: /Rdo 22) ROD: 

Therefore, the subtangent is, 


If now the polar equation of the curve RP be represented by the 
form, 


(173) R = gw, 
we deduce from (173) the value of 
dea 
dR’ 


and substitute it into (172). 


POLAR SUBNORMAL. 75 


The result is the length of the polar subtangent, involving tne 
co-ordinates of the point of tangency. (See Prop. F, App.) 
Ex.—The polar equation of the circloid (186f) is, 
(a) R = r cosa + 7, 
find the polar subtangent. The value of the differential coefficient 
deduced from (a) and put into (172), we have, for the subtangent 


x FIG, 41. 
R? 
(c) OD —_ — — -° 
7? SIN.w ees 
Since 7 sinuw = CS, the subtan- PR git 
. : oie : “ati 
gent of the circloid is a fourth propor- ae 4” 
tional to the three lines CS, AH, and ,/ \ 
« AG cs 
AH. Hence, the subtangent may be \ 
constructed, and the tangent to circloid x 
drawn. a .! z 
Ex, 2.—Find the polar subtangent in the curves. 
? 
R= —y R’= Ca*, 
1 + €. cos.e i 
R = logo, +» R? = ec cos.%0 + 67, 


PROPOSITION XXXIV. 


Determine the polar subnormal of a curve. 
FIG. 42. * 
If PM be drawn perpendicular to \ 


\ 


\ 


the tangent PD, and the. subtangent \ 

OD be produced, OM is the polar 

subnormal. M P 
Because in the right angled tri- 

angle MPD, PO is drawn perpendi- O 

cular to the hypotenuse MD, we R 

have PO, a mean proportional be- 

tween MO and OD, that is, 

(174) PO’ = OD.OM, 

substitute into this the value of OD, 

in (172), and we get, since PO = R, 

which is the length of the subnormal required. 


76 DIFFERENTIAL CALCULUS, 


We have, then, (178) from which to get the differential coeffi- 
cient used in (175). 

Ex. 1.—Find the polar subnormal of the circloid. 

The equation of this curve at (136/)) is, 
(a) R= cosisttt- ir. 

The value of dR ~ da, deduced from (a) and put into (175), 
we have, 
(5) OM = —7 sine. 

Hence, the subnormal equals CS, figure 41. 

Ex, 2.—Determine the polar subnormal in the curves H. 


me 


PROPOSITION XXXV. 


Find the length of the polar normal of a curve. 
The line PM, figure 42, is the polar normal of the curve PR. 
Putting N for the length of this normal, we have, from the right 
angled triangle POM. 
N? = PO + OM’. 
Put in this for PO, its equal R, for OM, its value in (175), and 
we have, 


(175a) N?— R? 4 oO 


which makes known the length of the normal. 

Ex.—Find the length of the normal in the circloid. 

Put into (175a) for dR ~ da, its value deduced from the equa- 
tion of the circloid, and we have, 
(175)) N? = 2r (t + f COS.a)'=="2'7R, 
which shows the normal to be a mean proportional between the 
diameter of the circle and the radius vector. 


PROPOSITION XXXVI. 


Determine the locus of the intersections of the tangent 
and polar subtangent in a plane curve. 


INTERSEC. OF TAN. AND SUBTAN, 
FIG. 43 


i a 


Let P be the pole, HR the curve, 
MH the tangent, and MP perpen- M 
dicular to PH, MP is the polar sub- 
tangent, and M is one point of 
the locus required, 


4 


. Pea) 7: 
Let (c,d) be the co-ordinates of P. Sic 


Let (x,y) be the co-ordinates of H, one point of tangency. 

Let (m,n) be the co-ordinates of M. 

Get the equation of HP, passing through the points H and P, and 
the equation of MP passing through P, and perpendicular to HP, is, 
for the point M, 


(177 ee Se ee AMT Pa) 
d—y 

The equation of the tangent line MH is, as in (93), for the point M, 
(178) n—y= p(m—2), 

The equation of the curve RH may be represented by 
(179) o(2z,y) = 0. 

At the point H, a and y are common to these three equations. 
Eliminate these co-ordinates between (177), (178), and (179). The 
resulting equation will contain no other variables than m and n, and 
may be represented by 
(180) o(mn) = o, 
which is the equation of the locus required. 

If either or both of the co-ordinates, (c,d), of P be zero, (177) 
rnust be modified accordingly, 

Ex.—Determine this locus when the curve is the ellipse, hyper- 
bola, or parabola, and the focus the pole. | 

Ans.—A straight line. 


LLIN AILS OL LLL 


% 


78 DIFFERENTIAL CALCULUS. 


CONSECUTIVE LINES AND POINTS. 


An idea of consecutive points of a curve may be formed by sup- 


posing the curve to be a polygon of an indefinite number of sides, 
AB, BC, &c. 


FIG, 44. 


The middle points S and E, of two adja- B 
cent sides of this polygon, would be conse- ~a-~S E 
cutive points on the curve, and are to be 
regarded as indefinitely near to each other. 

If through the two consecutive points S and E, normals be drawn, 
these normals would be consecutive lines; we may therefore say in 
general, that consecutive lines, (straight or curved), are lines inde- 
finitely near to each other, fulfilling given conditions. Thus in the 
case of the consecutive lines through 8 and E, the conditions to be 
fulfilled were, that they be normals to the curve. 

The following Proposition, and Examples under it, will elucidate 
the doctrine of Consecutive Lines. 


c* 


PROPOSITION XXXVII. 


A line, (straight or curved), is drawn subject to a given 
condition, determine the locus of the intersection of this 
line with its consecutive line. 

This general proposition may be more readily understood by ex- 
amining a few particular cases under it. Take first, the following. 
Ex, A. 

A straight line is drawn, cutting off, with the sides of a given 
angle, a triangle of given area, determine the locus of the intersec- 
tion of this line with its consecutive line. 


CONSECUTIVE LINES. 79 
PIG. 45. 


Let AX and AY be the sides ¥ 
of the given angle, and the 
axes of reference. Let CE be 
drawn, cutting off the triangle 
CAE of given area, This is 
the condition to which the line 


CE is subject. Let the equa- A aD E 
tion of CE be 
(181) y=—ar + B. 


As this equation contains two parameters, a and 8, let us seek, by 
the condition to which the line is subject, to eliminate one of them, 
For this purpose put s = area of CAE, and we have, from (181), for 


r=0, y=8=AC,andfor y=o, rx=>— == ARF. 
a 
Then for the area we have, 


(182) Psi sin. A 


a 2 

Substitute the value of a from (182) into (181), and we have, after 
putting m for sin.A ~— 2s, 

(183) y= — mer + 8B. 

This is the equation of CE, containing one parameter, 8, and con- 
ditioned to cut off a triangle, CAE, of a given area, As the para- 
meter § in (183), is arbitrary, it is obvious that the line CE, depending 
for its position on 6, may take an indefinite number of positions ful- 
filling the condition of cutting off the given area. Suppose BD be 
a line consecutive to EC, i.e., cutting off a triangley BAD = CAE, 
and since BD and CE are indefinitely near to each other, if we put 
BC = dg, and increase the g of (183) by ds, we have, 

(184) y = — mx (gp + dg) + 6+ df, 

which is the equation of BD, the line consecutive to CE. The point 
O, where these two lines intersect, is the point whose locus we are to 
find. 

It is obvious that the co-ordinates x and y, are common to (183), 
and (184), at the point of intersection O. Hence, if we subtract 
(183) from (184), and divide by ds, we have, 


.- 


a ee 
Eee tee 

2 ees Sree 

——sss a% 


ae 


a 


——— 


we fe 
we" 
ne 


f 


an 


. 


80 DIFFERENTIAL CALCULUS. 


(185) o = — 2m pr + 1 — madZ. 

If we suppose d@ indefinitely small, or zero, the term containing 
it in (185) may be omitted, and we have, 
(186) o = — 2m pr + 1. 

In (186) the co-ordinate 2 must still appertain to the point of in- 
tersection O. But when dé is indefinitely small, the lines (183) 
and (184) are what we have called consecutive lines, and the pro- 
cess by which we passed from (188) to (186), is obviously the pro- 
cess for differentiating (183) for # and y constant and g variable ; 
which is also made evident by the fact that if we differentiate (183) 
for B variable, and divide by dé, we have (186). Hence, if the 
position of a line depends upon a parameter, if we differentiate the 
equation of the line making that parameter the only variable, the 
co-ordinates 2,y of the differential equation will appertain to the 
point of intersection of the line with its consecutive line. 

Since at the point of intersection O the co-ordinates x,y are com- 
mon to (183) and (186), if we eliminate 6 between these two equa- 
tions, we have, 

1 


187 ry = ; 
( fi “ied 4m 


Equation (187) is the locus of the point O of intersection, that 
is, if the two consecutive lines CE and BD be imagined to move 
round within the given angle, subject to the condition that each 
always cuts off a given triangle, the locus of their intersection is 
equation (187). The curve (187) is tangent to the line CE in every 
position, for it may be conceived to be generated by a point O 
moving along CE as CE changes its position. (See Proposition H, 
Appendix. Cor.) 

From this process, we may infer generally, that if 
(188) (2Y,8) == T0; 
be the equation of a line, straight or curved, whose position depends 
upon the parameter 6; if we differentiate (188) for 6 variable, and 
x,y constant, the co-ordinates 2,y of the differential will appertain 
to the point of intersection of (188) with the consecutive line. Re- 
present the differential of (188) for 6 variable, and divided by 
dp, by 


CONSECUTIVE LINES. 81 


d.p(@,Ys8) __ 
(189) Becour 0. 

Eliminate 6 between (188) and (189), and the resulting equation 
in a and y, which may be represented by 
(190) F (x,y) = 0, 
will be the locus of the intersection of (188) with its consecutive 
line. ‘The line (190) is tangent to the line (188), for it may be 
conceived to be generated by the point of intersection (as O in fig. 
45,) moving along (188). 

(See Proposition E, Appendix.) 
We will illustrate this by several other examples. 


Ex. B. 
The centre of a circle moves along a given curve, find the locus 


of the intersection of the circle with its consecutive circle. 
FIG. 46. 


Let DR be the curve on y 
which the centre of the circle 
moves, and 
(191) p=, 
the equation of DR. 

Let P be the centre of the 
circle PH in one position. 

The equation of PH is, (a 


and @ being co-ordinates of P), oe 

(192) (7 — af + (y— 6) = RB, 
or substituting into (192) from (191), 

(193) (x —a)’ + (y — ga)? = R*. 


Here a is the parameter on which the position of the circle de- 
pends, and by the same reasoning employed in Example A, we could 
show, that if we differentiate (193) for x and y constant, and the 
parameter a variable, the co-ordinates 2 and y in the differential, 
which may be written, 

(194) (x — a) da + (y — 9a) d,pa = 9, 
will be the co-ordinates of the point E, where the circle PH inter- 
sects the consecutive circle QL. 

9 * 


82 DIFFERENTIAL CALCULUS. 


Eliminate a between (194) and (193), and the resulting equation, 
which will be of the form (190), will be the, locus of the point of 
intersection E. This locus will be tangent to the circle (193), in 
every position. 

To particularise the present example, suppose the line RD be a 
straight line, then (191) becomes 
(a) . 8 = ma. 

Substitute this into (192), and differentiate for a variable, and we 
have, 


(b) — 2 (x — a) da — 2m (y — ma) da = 0. 
Eliminate a between (b) and (192), and we have, 
(c) = mz + R(m? + 1)3 


which shows that the locus is two parallel lines, one of them being 
the locus of E, and the other of F. ' 

If the centre had moved along the axis of abscissas, (192) would 
have been, = 'O 
(d) (rx —ay?+ y= R’ 

Differentiate this for a variable, and eliminate a between (d), and 
this differential, and we have, 
(e) fae; 
which represents two parallel lines, whose distance apart is 2R. 

Instead of first eliminating 6 from (192), by means of (191), and 
then differentiating the result (193), we might first have differen- 
tiated (192) for a and 4 variable, and then have eliminated g and 
dg by means of (191) and its differential. This process, which we 
will have occasion hereafter to adopt, is the same as that shown in 
Proposition XXVIII., (w), (2c), &ec. 


Ex. C. 

The centre of a circle moves along a given straight line, its radius 
varies as the distance of the centre from a given point in that line; 
find the curve to which it is always tangent. 

Take the given line as the axis of abscissas. 


CONSECUTIVE LINES. 83 


FIG. 47 
Let the origin A be the given 
point, and a the distance from 
the origin to the centre P of the 
circle in one position. E 
Since the radius varies as the ii 
distance a, let Cie err wAN 
(d) R = ma, _ x 
be the radius where mis any 
numerical quantity. 
The equation of the circle PH is therefore, 
(e) (a—z)? + ¥ = ma’. 
Differentiate this for a variable, gives 
(f) a—xz= ma. 
Eliminate a between (f) and (e), and we have, 
g) aie penne 
(m?— 1)? 


an equation which shows that the circle is always touched by two 
straight lines passing through the origin. 
Ex. D. 

The centre of a circle moves along a given straight line, its radius 
varies as the square root of the distance of the centre from a fixed 
point on that line. Find the curve to which the circle is always 
tangent. 

Take the data as in Ex. C, and since the radius varies as the 
square root of the distance of the centre from the origin, we may 
express this law of variation by 


(e) R= J ma. 
and the equation of the circle is, 
(Ff) (a— zx) + 9? = ma. 


Eliminate a between (f) and its differential for a variable, and 
we have a parabola. 

We would proceed in the same manner if the radius varied as 
any given function of the distance of the centre from the origin. 
For putting, 

(g) R = 94, 


a3 
| 


\ 
c % 
a 
nL 
Te 
ae 
*h 


t 
bk 


; 


84 DIFFERENTIAL CALCULUS. 


the equation of the moving circle is, 
(h) (e — a) + yf = (pay. 

Eliminate a between (A) and its differential, for @ variable the 
result is the locus sought. 

Ex. E. 

The vertex of a parabola moves along a given curve, its axis 
remains parallel to itself. Find the curve touching the parabola in 
every position. 

This is readily done by imitating the steps in Ex, B. 

Ex, .F: 

A given angle moves along a given curve, one of its sides passes 
through a given point, find the curve to which the other side is 
always tangent. 


FIG. 48. 


Let PD be the given curve, ¥ 
CPR the given angle. Let the 
given point be R on the axis of 
abscissas. Let (a,@) be the co- 
ordinates of P, put AR = c 
and » = tan.CPR. xX 

The equation of PR passing through the two points P and R, is 

! Ca eg 
(k) Li Rie goin opersne! 

The equation of PC passing through P,and making the given 

angle with PR, is by Analytical Geometry. 


A 


v(a— 
(2) y—p= Catt eat 
Let now, 
(m) B = 94a, 


be the equation of PD. Eliminate 8 from (/) by means of (m), and 
then eliminate a between (7) and its differential for a variable, the 
result is the locus required. 

If the angle CPR, is a right angle, then » is infinite and (/) 
becomes, 


(7) y— p= 


which is the equation of PC in this case, 


Cla & (elt a), 


CONSECUTIVE LINES. 85 


To derive (n) from (7), divide the numerator and denominator of 
the fraction in (/) by v, and then putting v infinite (2) becomes (7). 

Ex.—If the curve PD becomes a straight line coinciding with the 
axis of y, then a = o and (n) becomes 


(0) y i 7. ic. 

Eliminate ¢ between (0) and its differential for 6 variable, and 
we have, 
( p) 2 = Ano 
a parabola, for the locus sought. 

When PD becomes a straight line coinciding with the axis of y, 
equation (/) becomes 


vc — B 
yY— p= ——. v 
(9) a igi 
which is the equation of PC. Find the curve to which PC is tan- 
gent in this case. Ans. A Parabola. 


What is the locus when PD isa circle, with centre at the origin, 
the angle CPR being a right angle. 
If 7 be the radius of the circle, the equation of the locus is, 
(r) 72 y? 4. y? (r? Pi c?) eile (7? ee c’) = 0, 
which is a circle when c is zero, an ellipse when c is less than r, a 
hyperbola when c is greater than 7, and a point when c is equal to r. 
This Proposition is the converse of Proposition XXIV. 


Ex. G: 
Between the sides of a right angle a straight line of given length 
is drawn, determine the curve to which this line is always tangent. 
Put m for the length of the given line and we find for the curve 
sought, the equation, 


"ie 2 
y + x = m’, 
Ex: H. 


An indefinite number of parallel lines are drawn to meet a given 
curve, where each parallel meets the curve, a line is drawn, making 
with the parallel, an angle which is bisected by the normal at that 
point; find the locus of the intersection of this line with its conse- 
cutive line. 


86 DIFFERENTIAL CALCULUS. 


FIG. 49. 


Let PD be the curve, 
SL one of the system of 
parallels meeting the curve 
InsP. 

Suppose PN the normal 
at P, and PG the line 
drawn making the angle N 
GPN == SPN. Draw BR tangent at P, and RE cael to lic 
axis of y. Let » equal the tangent of the given angle ELP, which 
each parallel makes with the axis of y. Pe 
(s) — as 
be the curve PD. 

The equation of the normal PN is, 


1 
(t) Aire aur See 
The equation of PG is of the form, 
(u) y— 8 = b (x —a), 


and it remains to express b in terms of constants, and of the co- 
ordinates of P. For this purpose, we have, 
(2a) tan.(ELP — LRP) = tan.LPR = tan.BPS = cot.GPN. 
(25) “. tan.GPN — se i > bane ARRAY a . 
tan.(ELP— LRP) ~~ po — 

Again, since b is the tangent of the angle PGN, we ie 
(2c) 6 = tan.PGN = tan.(PNX — GPN). 

Expand the last side of (2c), and by means of (2b), and recol- 
lecting that 

tan.PNX = —1 = p, 
we have, for b, the value, 
2po + p’—1 

peop — 1) 

Substitute this value of 6 into (wu), and we have for the equation 
of PG, 


(2d) b = 


2pv + p—1 
D) Sais! 6 ee, MIE OR cee 
te) a Sp v1) 


x— a). 


a 


CONSECUTIVE LINES. 87 


By means of (s), 8 may be eliminated from (2e), and the re- 
maining parameter a being eliminated between (2e) and its differen- 
tial for a variable, the resulting equation is the locus sought. 

Cor. Ist.—If the parallels SL be parallel to the axis of abscissas, 
AX, then v is infinite, and (2e) becomes 
2 ed eet pe 
(2f) 7 p= 1 —p (x a), 
which is, in this case, the equation of PG, 

Cor, 2d.—If the parallels SL be parallel to the axis of ordinates 
AY, then v is zero, and equation (2e) becomes 
(2g) y—p= Riss 

2p 
which is, in this case, the equation of PG, 

Proceed with (2¢), or (2f), as directed for (2e), and we have the 
locus sought. 

Example H comprehends the Theory of Caustic Curves, the rays 
of light being regarded as parallel, and the law of reflexion of light 
being that the incident and reflected rays make equal angles with the 
normal, at the point of incidence. 

For a detailed investigation of these curves, see the Analysis of 
the Marquis de L’Hopital, on the subject, in which they are examined 
with much neatness and elegance, and from more elementary con- 
siderations than those here employed. 

As an example, suppose the given curve be the parabola, 

y? = 4ma 
and the lines be parallel to the axis of abscissas, find the proposed 
locus. 

Using (2f), we have, for the curve required, 

y? + (x — my = 9, 
which designates a point, viz: the focus of the parabola. 

If instead of being parallel, the lines SP were drawn from a point, 
as at S, and made the same angle with the normal PN, which the 
normal made with PG, we could, by a similar process, obtain the 
equation of PG, involving as parameters the co-ordinates of P, and 
one of these parameters being eliminated from the equation of PG, 


(x — a), 


Hila 


88 DIFFERENTIAL CALCULUS. 


by means of the equation of the curve, we could, as before, find 
the curve to which PG is always tangent. This curve would be the 
caustic when the rays of light emerge from a point. 
. Bx. -K, 
A tangent is drawn to a given curve, at the point of tangency a 
line is drawn making a given angle with the tangent, determine the 
locus of the intersection of this line with its consecutive line, 


FIG. 50. 


Let PH be the given curve, rc 
and its equation 
(2h) 4) pe. 

The equation of the tan- 
gent line PR, is R A G 
(2k) y —y =p («' — 2a), 
where 2',y' are the co-ordinates of any point on PR. Let PG be 
the line drawn, making with PR, at the point P, a given angle RPG, 
Whose tangent we will represent by v. The equation of PG, pas- 
sing through the point P, is 


(2m) y —y = P42 (oo, 
1 — pv 
where « and y are the co-ordinates of the point of tangency, and 
x',y', the co-ordinates of any point on PG. By means of the curve 
(2h), eliminate one of the parameters x,y, from (2m), and we have, 
(2n) y — or = Buk Be (z' — x). 
1 — pv 
Putting in this the value of p, in terms of 2, deduced from (2h), 
we eliminate x between (2n), and the differential of (2n) for x varia- 
le. The resulting equation, which may be represented by 
(2p) g(t sy') = 
is the locus required. 
Suppose the given curve be the circle, 
a 4+ y? = R% 
Then proceeding as above, we have for the locus, 
c* a)" Soak COS,1er (x) 


RADIUS OF CURVATURE. 89 


Ex. L. 
A line of given length moves with its extremities along two given 
curves, determine the locus of its intersection with its consecutive 
line, 


Ex, M. 

A parabola passes through a given point, and has its vertex on a 
given curve, and its axis perpendicular to a given line, find the locus 
of its intersection with its consecutive parabola. 

Ex, N. 

A rectangle is constructed on the ordinate and abscissa of a given 
curve, a diagonal is drawn from the foot of the ordinate, find the 
locus of its intersection with its consecutive diagonal. 


PROPOSITION XXXVIII. 


Determine the radius of curvature at any point in a given 
curve. 


Definitions, 

1.—A circle which coincides with a curve at two consecutive 
points, is called, an Osculating Circle. 

2.—The radius of the osculating circle is called, the Radius of 
Curvature, 

3.—The point where the radius of curvature meets the curve is 
called, the Point of Osculation. 

4, The centre of the osculating circle is called the centre of cur- 
vature, 


Let EL be the curve, B 
and P consecutive points on 
it, and let the circle HBN 
coincide with EL at B and 
P, then C is the centre of 
curvature, CB the radius of 
curvature, and HBN is the 
osculating circle. 


FIG. 51. 


90 DIFFERENTIAL CALCULUS, 


In order to determine the radius of curvature at any point B of 
EL, proceed as follows. Let (a,) be the co-ordinates of any point 
on EL, and let the Equation of EL be represented by 
(195) 6 = 9a. 

Through the consecutive points B and P, draw BS and PR, nor- 
mals to EL, Since (a,8) are the co-ordinates of any point B on 
EL, the equation of the normal BS is, (see 67). 


(196) B—y + 2 (a—2)=0 


If we differentiate (196), for the parameters a and 6 variable, the 
x and y of the differential, as was shown in the previous proposi- 
tion, will be the co-ordinates of C the intersection of BS with its 
consecutive normal PR. The differential of (196) for a and g 
variable, a being the independent variable is, 


2 
(197) dp — da. “a (a —zr)+ __=0. 


The consecutive normals BS and PR intersect at the centre of 
the osculating circle. For since the curve EL and the circle BH 
coincide at P and B, the normals to the curve at B and FP are like- 
wise normals to the circle, and normals to the circle all pass through 
its centre. 

If x and y be the co-ordinates of the centre C, we have for the 
length of the radius of curvature BC, calling it R, 


(198) R= ((@—» + (@—2) 


At the centre C, the co-ordinates x and y are common to (196), 
(197) and (198). Hence, find the values of x and y in (196) and 
(197), and substitute them into (198), and we have the length of 
the radius of curvature in terms of the co-ordinates of the point of 
osculation B, and their differentials. 

The length of the radius of curvature found by eliminating x 
and y between these equations is, 


3 
dey? . ds slorpsrte a hake aig Lag 
(199) Rae (+4 Pa tap BSCE tare rane 


the last form being made from the first by putting, 


RADIUS OF CURVATURE. 91 
d ad? 

ap = 9 and _ — 7, 

da da’ 


This radius being a function of the co-ordinates of the point of 
osculation will depend on the nature of the curve EL, and vary 
when the point of osculation is varied. ‘To find it for any parti- 
cular curve (195), deduce the first and second differential coefh- 
cients from (195), and substitute their values in (199). 

Ex. 1. Let the equation of the curve EL be, 


(a) = ma", 
what is the radius of curvature? 
(b) Here orga 2ma = p and es tl oun a 
da da’ 
substitute these into (199), and we have, for the radius of curvature, 
3 

Am22\2 

(c) fate Ge ee ee 
2m 


Here the radius of curvature is a function of a, the abscissa of 
the point of osculation, and evidently increases as that abscissa in- 
If the abscissa a be zero (c) gives for the radius of cur- 


creases. 
vature at the origin. 
(d) R=2. 
2m 
Ex. 2. Find the radius of curvature in the following curves, 
£ = ma’, BA? = Ama, B = ma + na’, 


in which a is the abscissa and £ the ordinate. 

The radius of curvature (199) is positive or negative. ‘These 
signs only indicate the direction in which the radius is measured in 
respect of the curve. We have seen, Proposition XXX, that if the 
curve be concave towards the axis of abscissas, the second differ- 
ential coefficient is negative. In order, therefore, that the radius of 
curvature may be positive in such curve, (199) must be taken with 
the negative sign. Fora like reason, if the curve be convex to- 
wards the axis of abscissas, (199) must be taken with the positive 
sign, 
Since the osculating circle coincides with the given curve, at the 
puint of osculation, the curvature of the curve, at that point, is the 


be 


“<a 


OE | ESE 
A 


onl aad 


92 DIFFERENTIAL CALCULUS. 
same as the curvature of the osculating circle, and is therefore known, 
if we know the radius of curvature. 

Before leaving this Proposition, it may be proper to observe, that 
the radius of curvature (199) is deduced on the supposition that a is 
the independent variable, and consequently, (see Proposition XXIX.) 
da constant. Equation (196) was differentiated on this hypothesis. 
If, on the contrary, 8 be the independent variable, then dg is con- 
stant, and da variable, in differentiating (196), (see Proposition 
XXIX.) In order to obtain a form that will embrace both of these 
hypotheses, differentiate (196), as if neither a nor # were the inde- 
pendent variable, that is, consider da and dg both variable in (196). 
Then eliminate 2 and y from (198), by means of (196), and its differ- 
ential for both da and dg variables, and we have, for the radius of 


curvature, 


1 (de + dp’)? 
7" adeda — @adgp- 

If da be constant, or d?a = 0, (199a) becomes (199). 

If dg be constant, or d?3 = 0, (199a) shows the proper value of 
the radius of curvature on that hypothesis. If a and gp be both 
functions of some other variable quantity, taking that quantity as 
the independent variable, (199a@) shows the length of the radius of 
curvature. We will have occasion to employ this last principle in 
determining the radius of curvature of a polar curve. 


(1992) R= 


PROPOSITION XXXIX. 


Given a curve, determine its evolute. 
Definitions. 

1.—The locus of the centre of curvature is called, the Evolute 
of a Curve. Thus the locus of C, fig. 51, is the evolute of the 
curve EL. : 

2.—The given curve, when reference is had to the evolute, is 
called, the Involute. Thus EL, fig. 51, is the involute of which the 
locus of the centre C is the evolute. 
To determine the evolute, let EL be the given curve, and (195) 


EVOLUTE. 93 


its equation. We have seen, in Proposition XXXVIII., that the cen- 
tre C is the intersection of the normal (196), with its consecutive 
normal, Hence, eliminate the parameters a and g between the three 
equations, (195), (196), and (197), and the resulting equation, which 
may be represented by 
(200) (x,y) Oe 
is the evolute required. 

To perform this elimination most readily, first solve (196) and 
(197) for 2 and y, and we have 


(201) za —_Pp(i+p’) 
r 
(202) y= B+ pe ue 
Tr 


Substitute into (201) and (202) the values of the differential co- 
efficients deduced from (195), and then from these two resulting 
equations and (195), eliminate a and 8. 

This Proposition is evidently merely a corrollary to Ex. K, Prop- 
osition XXXVII. For when in Ex. K, the angle RPG becomes 
a right angle, PG becomes a normal, the locus becomes the evolute, 
and (2n) becomes, 


(202a) y —or = — hal (x' — x), 
from which, if x be eliminated, as in Proposition XXXVII., we have 


the evolute. 
Ex,.—Determine the evolute of the curves, 


2° = 4ma, the paraboia, 
as = m’, the hyperbola, 
mp? + n'a? = m?.n?, the ellipse. 


PROPOSITION XL. 


Normals to the involute are tangents to tne evolute. 

This is a corollary from the foregoing Propositions ; for the evo- 
lute being the locus of the intersection of consecutive normals, this 
locus, by the Theory of Consecutive Lines, in Proposition XXXVII. 

H 1G? 


94 DIFFERENTIAL CALCULUS. 


(see Proposition E., Appendix), is tangent to the normal, at its inter- 
section with its consecutive normal. Hence the truth of the Propo- 
sition. 


FIG, 52. 


From this it follows, that ifa ¥| & 
cord be wound round a given 
curve AP, and then unwound, the 
unwinding beginning at a point 
A, and the cord being kept tense, 
and in the plane of the curve AP, 
the extremity of the cord will 
describe the involute AM, of O 
which AP is the evolute. It is for this reason that the locus of the 
intersection of consecutive normals is called the evolute. 


PROPOSITION XLI. 


Determine the radius of curvature when the curve is re- 
ferred to polar co-ordinates. 


Suppose EL be the curve. 
Let the origin A, of rectangu- 
lar axes, be the pole, and let 
the axis of abscissas AX be 
the angularaxis. If (a,@) be 
the co-ordinates of any point 
B on EL, the rectangular 
equation of EL is, asin (195), A 
(203) B= 9a. 

Put p for the radius vector AB, and a for the variable angle BAX, 
then the values of a and @ in (203), are 
(204) a = pCOS.a, B= pSiNea. 

Here, since a and £ are both functions of p and w, we may take 
« as the independent variable, and obtain the first and second differ- 
entials of a and 8 in (204). Substitute these differentials into 


POLAR RADIUS OF CURVATURE. 95 


(199a), and we have the length of the radius of curvature in terms 
of polar co-ordinates. The result of this substitution is, 


9 a» 2 
~ u*) = 
(205) R= + 5 ( a 7) 
p + 2u* — pu 
d , ad? 
where u—- and u' — ual ? 
dus dix? 


We should have had the same result if instead of taking the axis 
of abscissas for the angular axis, we had taken any line inclined to 
that axis at an angle c. This is evident, inasmuch as the variable 
angle enters (205) only in the form of a differential. It may also 
be proved by taking instead, of (204), the values 

a@ = pC0S.(a — c), 6 = psin.(a — c). 

By substituting the differentials from these equations into (199), 
we have, as before, (205). 

Ex, 1. Find the radius of curvature in the circloid. 

Putting p for the radius vector, and m for the radius of the circle 
the equation of the circloid (136f ) is, 


(a) p = MCOS.w + mM. 
Differentiating this, we have, 
l ,; d? 
(b) Sr nina ad © PL — MCOS.w = u’. 


dis da 
Substitute the differential coefficients (5) into (205), and reducing 
by (a), we have for the radius of curvature R, the value, 


1 1 
(c) R = 2 (2m? + 2m’cos.a)? = 2 (2mp)?, 

Comparing (c) with (175), we see that the radius of curvature 
in the circloid, is two-thirds of the polar normal, 


Ex, 2. Find the radius of curvature of the curves, 


§P =! Cis", p” = c’cos.20 +. 6’. 
(K) | at 
é p= m 102.0, p=". 
1 + € COSe. 


PROPOSITION XLII. 
Lemma. 
In any equation of the form, 
(206) y' = o(x +h), 


96 DIFFERENTIAL CALCULUS. 


the differetial coefficient is the same for x constant and A va- 
riable, as for A constant and x variable. 


FIG. 54. 


If we put in the curve AH, + 
(A) AE=2, PE=y, 
ES = h, Be Y's 
equation (206) may be re- 
garded as the equation of the 
curve AH, and it is evident in 
the curve that y' varies forh a 
constant and x variable in the E S 
same manner and degree that it does for x constant and h variable. 
This would give us 


(206a) aes a 
dx dh 

As a particular case, let (206) be, 
(a) y = (x + hk)”. 

Differentiate this for / constant and a variable, and we have, 
(b) a —n(e + hy 

Differentiate (a) for x constant, and h variable, and we have, 
(c) a = n(x hr. 


The value of the differential coefficients (b) and (c), are the same. 


PROPOSITION XLIII. 
Taylor's Theorem. 

Determine a general method of expanding into a series, a 
function of the sum or difference of two independent va- 
riables, 

Let us represent the equation of the curve AH by 


(207) y = on. 
If x be increased by h, the equation of the curve is, 
(208) y =o (x + h), 


in which h may be represented by ES, and y' by HS. 


TAYLOR’S THEOREM. 97 


The object of the proposition is to expand the second side of 
(208). For this purpose let o(2 + h) be assumed equal to a series 
in the ascending powers of Ah, viz. 

(209) o(e@ +h) = A+ Bh + Ch? + DAF 4+ &e,, 

or, which is the same, put 

(209a) y =A+ Bh+ Ch? + DA? + &e.,, 

where A, B, C, &c., are indeterminate coefficients containing 2 but 
not h, 

It remains to determine the coefficients A, B, C, &c., and their 
values being put into (209), we will have the expansion of 9(a + h). 

To determine these coefficients, we observe, that the first term of 
(209) being independent of hk will remain unchanged, whatever be 
the value of A, and consequently, will not change though h = o. 

But when h = 0, (208), and (209a), become (207), consequently 
we have from (209), or (209a), 

(210) Ae = y = 2. 

To determine the other coefficients, B, C, &c., differentiate (209a) 
for h variable, and 2, (i. e. A, B, C, &c.) constant, and then differ- 
entiate (209) for & constant and x variable, and since by (206a) of 
the preceding theorem, we have, 

ES ii csr AY" 

dh © ida? 
we will have, by equating the values of these differential coefficients 
from (209a), 


(211) B+4+2Ch-+ 3Dh? + &¢,= ee ee dC 
x 


dB 
— +h’, 
dx * dx 

By the principle of the method of Indeterminate Coefficients, the 
coefficients of the like powers of h, on each side of (211), may be 


put equal to each other. This furnishes the equations, 


- &c, 


(212) Been 
dx 
(213) cng ae: 
dx 
dC 


(214) 3D = —, &e. 
dx 


98 DIFFERENTIAL CALCULUS. 


By means of (210), equation (212) becomes, 


dy d’y 
215 = er 6b = eee 
(215) - hen ioead 
and (213) becomes, 
; ey ay vs 
216 = ° Th dC —  ——=—. 3 
( ) Qda? a 2d? 
and (214) becomes, 
(217) oe feeccl: Wabh 

2.3.dx° 


Substitute the values of A, B, C, D, &c., from (210), (215), 
(216), (217), into (209a), and we have for the development required, 
OL; aprena dy 724, ay 
cD Posntestvggs Bay Mie popasi e chsacsr as 

This serves to determine the ordinate y' in terms of any other 
ordinate y, of the increment /, and of the differential coefficients of 
y. Inthis theorem, A is not supposed indefinitely small, as in the 
procedure in Proposition I., but is of any value whatever. 

Transpose y in (218), and we have, 
= font ee iy dy 49 
Mince 8 ade sig Raa OR UES : 

The second side of (219) shows the increment HC, in terms of 
h, and of the co-ordinates of P. 

Comparing (207), (208), and (218), the latter equation may be 
written, 


h® + &c. 


: aid dox Goer ss | Pots, ~ee 
(220) g(r +h) = ox + > oa WO saat -- apts! h3 + &e. 
which shows, at a single view, the development of p(x + h). 

This theorem is of very extensive application, and is employed by 
most writers on the Calculus, in resolving the Geometrical Proposi- 
tions that have occupied our attention in the previous part of this 
work. We will have occasion to employ it hereafter. 

As an application of this development, let us develope 
(2) Yee hye 

This is a particular case of (208). By making h zero in (a), 
we have, 


MACLAURIN ’S THEOREM. 99 


(4) y= 2m, 
which is what (207) becomes in this particular example. 

Deduce from (6), the successive differential coefficients, and sub- 
stitute them into (218), and we have, 


(c) y = (@ + hp = 24+ nah + net) ae xh? + &c. 


In the same manner, develope 
Sie log. (x + h), y = log.(« —h). 
y = sin.(a« + h), y = tan(2 + h), 

If equation (208) had been, 

y = o(x a h), 

the algebraic signs of (218), (220), would have been alternately 
positive and negative, i.e, the odd powers of h in (218) or (220), 
are negative, 


PROPOSITION XLIV. 
Maclaurin’s Theorem. 


Determine a general method of expanding into a series a 
function of a single variable. 

Suppose we have the equation, 

(221) Y = 9%, 
which may or may not contain constants. 

Assume the second side of (221) equal to a series in the ascend- 
ing powers of 2, containing the indeterminate co-efficients A, B, C, 
&c., that is, 

(222) y= A+ Be + Ca? + Da? + &e. 

The second side of (222) is the development of oa, and it re- 
mains to ascertain the values of A, B, C, &c. 

To determine these coefficients, we observe, that (222) being, by 
hypothesis, the true development, whatever be z, it is therefore, true 
when x = 0. If we denote by y’ the value of (222), when x = 9, 
we have, 

(4) y =A. 

Differentiate (222), and we have, 


(0) oe = B+ 2Cx + 3Da? + &e. 
v 


100 DIFFERENTIAL CALCULUS. 


This, also, must be true when x = 0. Whena = 0, (b) becomes 


dy’ dy’ 
c =~ == ws, Where by,—~; 
(¢) dz' "Dt bea dx' 
we denote the value of ee when 2 1s zero. 
Ey 
In the same way differentiate (b), and put 2 = 0, and we have, 
(d) oe Cr eS 
it ee 2dx'? 
Differentiate (b) twice and put 2 = 0, and we get, 
3 / 
() Digohicay 
2.3.dx'” 


Substitute these values, (a), (c), (d), (e) into (222), and we have, 
226 = y' dy’ ay 2 Dy’ 13 
OS Bie Satins tase ROT sagt” tapas” TO 
In (228) the accented differential coefficients and y' are used to 
denote the value of the differential coefficients and original function 
of x when x = 0, 
As an application of this formula, take the following example. 
Ex. 1.—Develope cos. into a series in terms of 2. Put 


(f) hu OSC, 

and when 2 = 9, COS. ==" Tadius, OF y= E: 
From (f') by differentiating, we have, 

(g) wo = — sin.x, and when x = 0, we have, ay rr, Oe 
Differentiate (g), and we have, | 

pp ahead ay’ 

(h) aa = — cos.z, and when x = 0, we have, 8 =—], 


and so on, the odd differential coefficient being zero when zr = 0, 
and the even one plus and minus unity alternately. These values 
substituted into (223), we have, 
bad x' x® 
(k) Us— COs.e =") of are ae he 
Equation (k) shows the length of the cosine of an arc in terms 
of the radius (unity) and of the arc a, 


After the same manner develope, 


TANGENT PLANE. 101 


y = sin.2, y= tan.2, y = cos.7'2, 
y == tan."2; Yo a= sin. 'a. 


PROPOSITION XLV. 


Determine the equation of a plane tangent at a given 
point, on a given curve surface. 
The equation of a curve surface contains three variables, and 
may be represented generally by the form, 
(224) 9(2,y,%) = 0, 
which denotes an equation containing three variables. 
Particular forms of (224) are 
e+-yt+vwtrao, the sphere. 
(L) 2? + y? — pz = 0, the paraboloid. 
A??? + Ba? + By? — A®B? = o, the ellipsoid. 


If the equation of a surface be solved for one variable, as 2, it 


will be of the form, 
(225) 2 = 9(2,y). 
We will use either of the forms, (224), (225), according to con- 


venience. 


FIG. 53, 


Suppose (224) be 
the given surface, P the 
given point on it, and 
(x,y,z) the co-ordinates 
of P. 

If 2',y’,z', be the 
variable co-ordinates 
on a plane, then the 
equation of any plane 
DCB, passing through 
the point P, is of the 
form 
(226) Z’— % = mx’ — x) + n(y'—y), 
where 


m = tan.DCX, n = tan.DBY, 
the angles which the traces of the plane, (226), make with the axes, 
1] 


102 DIFFERENTIAL CALCULUS. 


Suppose plane (226) tangent to surface (224) at P, and that DC 
and DB are its traces, 

Through P pass an auxiliary plane parallel to the plane ZY. 

This auxiliary plane cuts plane (226), in the right line EF, par- 
allel to the trace DB. 

The same auxiliary plane cuts the curve surface (224), making a 
section whose plane is the auxiliary plane. 

This section will have the right line EF for its tangent at P. The 
traces EG and GII, of the auxiliary plane, are parallel to the axes 
AZ, AY. 

The tangent line EF makes with GH an angle EFH, equal to 
DBY. 

The equation of the section made by plane EGH, is found by 
making, in (224), a constant, and equal to AG. 

The differential coefficient i deduced from the equation of this 

y 
section, expresses the tangent of the angle KFH, (Proposition II.) 
Therefore we have, 


(227) eae n. 


In the same manner, by passing a plane through P, parallel to 

the plane ZX, we would have, 
dz 

(228) as agit 

The differential coefficient (227), is supposed to be deduced from 
the equation of the section made by plane EFH. But as this section 
is made from surface (224), by making x constant, we get (227) 
directly, by differentiating (224) for x constant. 

In like manner, we get (228) by differentiating (224), supposing 
y constant. 

Substitute (227) and (228) into (226), and we have for the equa- 
tion of (226), 


: dz,, ga 
299 yt) — —F —— i 13 ee —— é 
(229) oe) a (y' —y) 


This plane contains two lines at right angles, tangent to surface 


TANGENT PLANE. 103 


24). Equation (229) is therefore the tangent plane to surface 
24 “ at the point (7,y,%)- 

To apply (229) to a given surface, deduce from the proposed sur- 
face (224), the differential coefficients, as above directed, and substi- 
tute their values into (229), 

Ex.—Find the tangent plane to the sphere. 

In this case, (224) becomes, 


(2 
(2 


(a) er+y+2—r=ao. 
Differentiating (a) first for x constant and then for y constant, we 
have, 
(5) Pie Fang a 
dy % x % 


substitute these values into (229), and subtracting (a) from the re- 
sult, we have, for the plane required, 
(c) re’ + yy + 27,’ —r=—o. 

Ex. 2. Find the tangent plane to the surfaces, 

x? + y? — 4mz = 0, a’y? + a’a? + b'2? — a’bh? = o. 

From the foregoing we observe, that when one of the co-ordinates, 
as x, in the equation of a surface, is constant, and the other two 
variable, the variation, differentiation, &c., takes place in the section 
made by a plane parallel to the co-ordinate plane ZY. Similarly 
when y is constant, and x and z variable, the variation, differentia- 
tion, &c., takes place in a section parallel tothe plane ZX. For 
these reasons, the equation of a surface is an equation containing two 
independent variables. 

In the foregoing and subsequent propositions, we will consider x 
and y as the independent variables, z being the dependent variable. 


The differential coefficient deduced from the equation of a surface, 


on the supposition that one of the co-ordinates is constant, is called, 
The Partial Differential Coefficient. Thus os and vi in (b), are 
E- ly 
partial differential coefficients deduced from the sphere (a). 
Let the sections of a curve surface, which are made by planes 
parallel to the co-ordinate planes, ZX and ZY, be called, Parallel 
Sections, and let the traces on XY, made by the planes of the par- 


allel sections, be called, Parallel Traces. The section of the sur- 


104 DIFFERENTIAL CALCULUS. | 


face (224), made by the plane EGH, is a parallel section, and GH 
is a parallel trace. 

From (227), (228), we observe that the partial differential coeffi- 
cient of a curve surface expresses the tangent of the angle which 
the tangent line to a parallel section makes with its parallel trace. 

When the partial differential coefficients (227), (228), exist to- 
gether, it is obvious that the tangent lines are drawn from the inter- 
section of the parallel sections, and lie in the plane tangent to the 
surface at that point. 


PROPOSITION XLVI. 


Determine the point on a given curve surface through 
which, if a tangent plane be drawn, its traces will make 
given angles with the co-ordinate axes. 

Let (224) represent the given surface. 

Put m and n for the tangents of the given angles, and we have 
(224), (227), and (228), three equations to determine a, y, and <, 
the co-ordinates of the point required, 

Ex. 1,—Determine such a point on the surface of a sphere. 

Here (224) becomes, 


Lz 4 dz Ka 
BNP 8 PG We SS ei etl nee eae 
i ai iC pe helpers and o 
and (227), (228), become, 
(b) ee en and) ene ay. 
% z 


Solve the two equations (b), and equation (a), for x, y, and z, 
and we have, 
. — nr — mr 
ae mee 2 13 and & = ae 
(m? + n? + 1)’ (m+ n* + 1)? (m?+ n?+-1)* 
for the three co-ordinates required. 
Ex. 2.—Determine such a point on the surfaces, 


y? + a? — 4mz = 0, Cyt evr’ +Pr—as’=o, 


MAXIMUM ORDINATE OF A SURFACE. 105 


PROPOSITION XLVII. 


Determine the point on a given curve surface at which the 


ordinate z is a maximum or minimum. 
FIG, 54. 


Let the equation of 
the given surface be, 
(230) 9(2,y,7) = 9. 

Through the point 
required, suppose a 
plane tangent to the 
given surface be drawn. 

The traces of a tan- 
gent plane make an- 
gles with the axes of X 
and Y, whose tangents 
are the partial differen. /¥ 
tial coefficients, (227), (228). But when the ordinate z of the 
point of tangency is a maximum or minimum, the tangent plane 
must be parallel to the plane XY. In that case the points B and 
C are infinitely distant, and the angles DCX and DBY are zero or 
180°, and their tangents are zero. Hence, when z is a maximum 
or minimum, (227) and (228), become 


dz 
231 = 
(231) n= 
lz 
(232 Crean 
(232) =e 


Equations (230), (231), and (232), make known the co-ordinates 
of the point on the surface whose ordinate is a maximum or 
minimum, 

Ex, 1. Determine the maximum or minimum ordinate when the 
surface is a sphere. 

If (a,3,c), be the centre of the sphere, its equation is, 

(a) (x — a) + (y— By + (2 —cy—R =o. 


By differentiating (a), we have, 
11 * 


DIFFERENTIAL CALCULUS. 


it (c) Pile AN as and oo ee fa 
x dy Z—ec dz z—c 
These differential coefficients being as in (231), (232), put equal 
to zero, we have, 
(d) = eg eon x 
Z——€ b-—4e 
We have the three equations (a) and (d) to solve for 2, y and z. 
These give, 
(e) - fee y = 6, and = = ¢ oc ix 
which are the co-ordinates of the point on (a), whose ordinate z is 
& maximum or minimum. 


Ex. 2. Determine the maximum or minimum ordinate in the 


surface, 
(Tf) 2 = ary — xy — zy’. 
Differentiating this first for 2, and then for y constant, we have, 
dz ‘ Ly 
(g) ia = ar — 2 — 2y. 
(h) 3 = ay — 2xy — y’. 
dx 
Put (g) and (hk), each equal to zero, as in (231) and (232), and 
we have, 
(k) ax — x? — 2xry = o, and 
(2) ay — 22y— y = 0. 


Solve (%) and (2) for y and 2, and we have, 


a 


(m) Pe t= 0, 
’ — a . — 
(n) Ste 
ia To distinguish the maxima from the minima ordinates in cases 


where the nature of the proposition does not immediately make 
known which is a maximum and which a minimum, we have the fol- 
lowing considerations. 

When the surface is concave towards the plane XY, it may have 
& maximum ordinate z drawn from some point on its concave sur- 
face, but not a minimum ordinate. If it be convex towards the 
plane of XY it may have a minimum ordinate drawn from some 
point on the convex surface to the plane XY, but not a maximum 


MAXIMUM ORDINATE OF A SURFACE, 107 


ordinate, But if a surface at any point be concave towards the 
plane of XY, parallel sections at that point are concave towards the 
parallel traces on XY, and conversely if the surface be convex. 
Hence, by the reasoning in Proposition XXXI, when the surface has 
&@ maximum ordinate, we have, 


dz 

—. = a negative quantity, and 
dy? 

d’z ; 

—— = a negative quantity, 

dx? q 


and conversely when the surface has a minimum ordinate, 
To apply this test to Example 1, differentiate equations (c), and 
by means of the values in (e), to wit: 


(0) x= @, y = B,andz = c + R, we have, 
(P) OM eee ean Chee Ls 
dy? R dx” R 
By means of the negative value of z in (e) to wit: 
(q) oe aor th, y = B,andz =c—R, 
the second differential coefficients of (c), become 
(r) RO ta tant tt en oe 
dy’ R dx? R 


Hence, the value of the co-ordinates (0) give a maximum, and 
the value of the co-ordinates (q) a minimum ordinate z. 

The same values of x and y that render z a maximum or mini- 
mum in the expression, 
(233) z= ory), 
will render it a maximum or minimum whether (283) be the equa- 
tion of a surface, or have any other signification. 

This obvious principle extends this Proposition to an interesting 
class of Problems. As an example, take the following. 

Ex, 3.—Divide a given straight line into three parts, so that their 
product will be the greatest possible. 

Let a = the line, and let x,y, and a — + — y be the three parts. 

Then putting z for the continued product, we have, 
(s) z= ary — wy— xy’, 
the same as (f ), above, and the values of the parts z and y are de- 
termined as in the process (f') to (n). 


108 DIFFERENTIAL CALCULUS. 


The value of x and y in (m) and (n), show the line to be trisected. 

Ex. 4. Divide a given straight line into three parts, such that the 
product of one part, into the square of the second, into the cube of 
the third, will be a maximum. 

Put z for the product, a for the line, z,y, and a — x — y, for the 
parts, and we have, by the enunciation, 

z= «ya — 2 —y). 

Clear of the vinculum, and proceed as in (f') to (n). 

Before leaving this Proposition, we will explain a process often 
found convenient in practice. 

Resume the example. Divide a given straight line into three 
parts, such that their product will be a maximum. 

Put a = the line, and let a, y, and wu be the three parts, and z = 
the product. 

Then by the enunciation, we have, 

(2a) x+y+u=a,and 
(2b) gine Neath, 

Instead of eliminating u from (2b), by means of (2a), differen- 
tiate (2a), and (26), for x,y and wu variable, and we have, 
(2c) dx + dy + du=o. 

(2d) dz = xydu + xudy + uydz. 

Eliminate wu and du from (2d), by means of (2a) and (2c), and 
we have, 

(2e) dz = axdy — x’dy — 2xrydy + aydx — 2xydx — y'dz, 

Put the terms on the second side of (2e), that contain dy equal to 
zero, and the terms that contain da equal to zero, and we have the 
equations 
(2f,) ax — x’ — 2xry = 0, 

(2g) ay — 2ry — y’ = 0. 

These equations are the same as (k) and (J). 

This process might be explained generaily, but the foregoing will 
suffice. Illustrative of the process, take the following example. 

Ex. 5, Given any number of plane areas situated on given 
planes, determine the plane on which, if all the given plane areas be 
orthogonally projected, the sum of the areas of the projections will 
be the greatest possible. 


NORMAL TO A SURFACE. 109 


The planes of the given areas being known in reference to three 
co-ordinate planes, let the given areas be each projected on each of 
the co-ordinate planes, Let 


M = the sum of their projections on the plane XY, 
M’ = 6 éé $6 XZ, 
MM" — ‘“ és Ts VAS 


Let x, y, and u be the angles which the plane required makes with 
the co-ordinate planes XY, XZ and ZY respectively. If z be put for 
the sum of the projections of the given areas on the plane required, 
we have, as is shown in books on Analytical Geometry, 


(2h) * = M cos.2 + M’ cos.y + M” cos.w, 
By Analytical Geometry, we also have, 
(2k) cos.*x + cos.2y + cos2u = 1. 


Proceed with (2h), and (2%), as was done with (2a), (26), and 
we have for the required angles. 
M M' 
= ——~ J. cos.y = a 
(M? + M? + M"2)2 (M? + M? + M’)2? 
(M? + M? M2 
The three angles thus determined make known the position of 
the required plane. 
This is the plane of Greatest Prajection, and is the same that is 
known in Mechanics by the name of The Invariable Plane. 
Ex. 6. Given the solidity of a rectangular parallelopipedon, de- 
termine the edges when the surface is a minimum, 
Ans. The edges are equal. 
Ex. 7. Determine the three angles of a plane triangle, so that 
the product or sum of the sines of the angles will be a maximum.. 
Ans. The angles are equal. 


COS.V 


CUS == 


PROPOSITION XLVIII. 


Determine the equation of a normal line at a given: point 
on a curve surface, 
I 


410 DIFFERENTIAL CALCULUS, 


FIG. 55. 


Let the equation of the 
surface be 
(234) o(a,y,%) = 0 

Suppose P the point on 
the surface through which 
the normal is to be drawn. 
Through P pass a plane 
tangent to the surface. 

Let DC and DB be the 
traces of such a tangent 
plane. 

If 2’, y', and z’, be the variable co-ordinates of this plane, its 
equation is, (see 229). 


(235) Zz —s=p (x —xr)+qy— y)s 
where we have put, for brevity, 

A Heer lz 1% 

236 at Pd  Gisty — 
(236) Ries. tay 


abbreviations frequently used hereafter. 

Since the normal to the surface is perpendicular to the tangent 
plane (235), the projections HS and RL of the normal are perpendi- 
cular to the traces DB, DC of plane (235). ‘The normal passing 
through the point x,y,z on the surface, if x',y' and 2’, be the va- 
riable co-ordinates on the normal, the equations of its projections 


HS, and RL, are, 

ts §z’—z=a(a' —2), equation of HS, 

(280) dz’ —2z = b(y' —y), equation of RL. 

Recollecting that p = tan.DBX, and q = tan.DCY, we have by 
1 


the usual relation of the tangents of perpendicular lines a = — _, 


1 P 
and b = — —- These values change (237) to, 
u 
238 HAC iD Rare a Be 
( ) 5 eli UL) jae & 


These are the equations of the normal line required, in which p 
and q are the partial differential coefficients from the surface (234). 
Substitute into (238) for p and q, their values deduced from the sur- 
face (234), and we have the equations of the normal line. 


LOCUS OF THE POINT OF TANGENCY. 111 


Ex.1. Find the equations of the normal line to the sphere at the 


point x,y,z. 
Here (234) becomes 


(a) e+ y + 2p? — o, 

from which we have, 
dz x z r 
= = =p and St = Gg, 
dx % dy z 


and (238) becomes 
, x ; 4 
(C) 2 —ax = — (2'—xz)andy —y= 7 (z' — 2). 
Equations (c) are the equations of the normal required. 
Ex, 2. Find the equation of the normal to the surfaces 
x 4+ y® —4Amz = 0, a’y? + bx? + 2 — mn? — o. 
a® y? 4. ba? + 27) — a? B=o0, a? y — (2? + 2?) + a Bo, 


PROPOSITION XLIX, 


A tangent plane is drawn to a given curve surface, its 
trace on one of the co-ordinate planes makes with the co- 
ordinate axes in that plane, a triangle of given area, deter- 
mine the point of tangency. 


Let DBC be the tangent 
plane, 

Suppose the plane on 
which the triangle of given 
area is made, be the plane 
of XY. Let the surface 
be, 

(239) o(2,y,2) = 0. 

The equation of the tan- 
gent plane is (235), 


In (235), for z’ — 0, x’ = 0, we have, 
(240) y' = PT + ay — alg hy 3h 


q 


112 DIFFERENTIAL CALCULUS. 


In (285), for z' = 0, y' = 0, we have, 
(241) et DE PY ore 


4 

Put v? = area CAB, and we have, from (240) and (241), by the 

rule for the area of a triangle, 

(242) 2pqv’ = (px + qy — 2)’; 

in which the partial differential coefficients p and q, are to be deduced 
from the equation of the surface (239). 

Equations (239) and (242) are all the equations we have for de- 
termining the point of tangency. But as these two equations con- 
tain three unknown quantities, z, y, and z, there are an indefinite 
number of points, (x,y,z), through which the tangent plane. may be 
drawn. If we first eliminate one of the variables, as z, between 
(239) and (242), and represent the resulting equation by 
(243) FE (2,4) =;0, 
and then eliminate another of the variables, as x, between (289) and 
(242), and represent the resulting equation by 
(244) 1(2,y) = 0, 
the two equations, (243) and (244), will be the projections on the 
planes of XY and ZY, respectively, of the line on surface (239), 
which is the locus of the point of tangency. There may, therefore, 
be an indefinite number of tangent planes to surface (239), each 
cutting off the proposed triangle, the points of tangency forming on 
surface (239), a curve whose projections are (243), (244). If one 
of the co-ordinates be given,as 
(245) o.= ™, 
we have three equations, (239), (242), (245), to find x, y, and =, 
which being determined by the solution of these three equations, 
make known the point of tangency. 

If we put into (235) the values of p and qg, taken from the sur- 
face (239), we may employ the resulting equation of the tangent 
plane in forming the equation (242), 

Let the surface be the sphere with centre at the origin. 

Here (239) becomes, 


(a) ety + 2—R =o, 


POINT OF TANGENCY. 118 


and the equation of the tangent plane to the sphere is, 
(b) ru’ + yy’ + xz’ = R% 

Proceed with (b) as in (240), (241), and we have, for.the area 
corresponding to (242), 
(c) 20°ry = R‘. 
| Equation (c) shows, without further elimination, that the projec- 
tion on the plane of XY, of the locus of the point of tangency on 
the sphere, is a hyperbola, whose asymptotes are the axes of X and Y. 

Eliminate x between (a) and (c), and we have a line of the fourth 
order, for the projection on ZY of the locus of tangency. If we 
suppose one co-ordinate z given, as in (245), we may find the co- 
ordinates of the point of tangency, by solving (245), (a), and (ec), 
for x, y, and z. 

Ex. 2,—Determine the locus of such a point of tangency in the 
surfaces, 

a? + y? — 4mzr,=— 0,a? x? + a? y? + b? 2? — a? B= 0. 


PROPOSITION L. 


A tangent plane is drawn to a given curve surface, its 
traces on two of the co-ordinate planes make with the axes 
triangles, each of whose areas is given, determine the point 
of tangency. 


FIG. 57. 


Let CAB and DAB be 
the triangles of given 
area. 

Put u? = the area of 
DAB. Proceed as in last 
Proposition, to (242). 

Then to find AD, put 


xr = 0, y = 0, 
in (235), and we have YY 
(246) 2'= 3 — (pe + qy) = AD. 


i a eS a on 
> . Tae x Pa 


a an 
- Pen 
Pe es 


Seah ns Se 


114 DIFFERENTIAL CALCULUS. 


By means of (246), (241), we form the expression for the area 
DAB. Hence, 
(247) 2pu> = — (px + qy — 2) 

From equations (247), (242), (239), we can find 2, y, and z, the 
co-ordinates of the point of tangency required. 

Ex. 1.—Let the surface be the sphere, find such a point of tan- 
gency. 

Ex, 2.—Let the surface be the paraboloid or ellipsoid of revolu- 
tion. Find such a point of tangency. 


PROPOSITION LI. 


Determine the distance from a given point on a curve 
surface to a given plane, measured on the normal to the 
surface. 

Call P the point on the curve surface, and D the point where the 
normal meets the given plane. Let 


(248) Az' + By’ + Cz' + D= oa, 
be the given plane, and 
(249) 9(x,y,2) = 9, 


the given surface. 

If N be the distance PD, the length required, we have from Ana- 
lytical Geometry, (putting (a,y,z) for the co-ordinates of P, and 
(x',y',z') for the co-ordinates of D). 

(250) N= (( ae ee ee) a a Coe yy 

The equations of the normal line are, by (238), 

(251) Ne —-%)+ “2 —zx=0. 
q(z —z)+y—y =o. 

At the point where the normal line pierces the plane, the co-ordi- 
nates x’, y’, and z’, are common to (248), (259) and (251). Find 
these co-ordinates from the three equations (248), (251), substitute 


into (250), and we have the length required. 
Ex.—Find such a distance in the surfaces, 


POINT OF TANGENCY. 115 


(M) oe Be a ee sie, r+ y—~ 4mz = 0, 


a(e+ xy) + OP—a@vP=o, 2+y—m 


PROPOSITION LII. 


Determine the distance from a given point on a curve sur- 
face to each of the co-ordinate planes, measured on the nor- 
mal to the surface. 

Where the normal pierces the plane of XY, the ordinate z' is 
zero in (250) and (251), and these equations become 


(252) Natl a (Qarreo it) iors (MS eh cha ): 
(253) te ct oe 
se 45) oe fm BY ee 


Eliminate 2’ ,y', from (252), by means of (253), and we have for 
the distance required from the plane XY, 
(254) N = 2(p’'+ q@ + 1)?. 

Proceeding in the same manner, we find for N' the distance re- 
quired from the plane ZY, 


(255) N= - Cp pega 17! 


of— 


and for the distance N"’ from ZX, we find, 
(256) Noes & (P+ ¢ +1), 


Ex.—Determine these distances in the surfaces (M), last Propo- 
sition, 


PROPOSITION LIII. 
A plane is drawn through a given point, and tangent to 
a given surface, determine the point of tangency. 
Let (a,b,c) be the co-ordinates of the given point, and (x,y,%) the 
co-ordinates of the point of tangency. 
The equation of the tangent plane is (235), 
(257) 2 —%t=p(«e —zr)+q(y —y). 


= ree Ate 


a ee 
bn sae ee 


ae a acl se > =n mepcaampenein peuncaoee 
Se < 


116 DIFFERENTIAL CALCULUS. 


t the given point, the co-ordinates 2’, y', and z’, of this plane 
become a, 6, and c, respectively, and (257) is, 


(258) c—z=p(a—zr)+q(b—y). 
Let now the equation of the surface be 
(259) O(2,Y,%) = 0. 


At the point of tangency, the co-ordinates x,y,z are common to 
(258), (259), and these are the only two equations we have to de- 
termine the point of tangency. If we first eliminate one of the var- 
iables, as z, between (258) and (259), and represent the resulting 
equation by 
(260) F (#,y) = 0, 
and then eliminate another of the variables as 2, between (258) and 
(259), and represent the resulting equation by 
(261) Yay) = 0, 
the two equations (260), (261), will be the projections on the planes 
XY and ZY, of the line on surface (259), which is the locus of the 
point of tangency. There may therefore be an indefinite number of 
tangent planes to surface (259), each passing through the given point, 
and the points of tangency forming on surface (259) a curve whose 
projections are (260), (261), If one of the co-ordinates, as z, of 
the point of tangency be given, we can determine the other two from 
(259) and (258). 

Ex. 1.—Let tne given surface be a sphere. Then (259) is, 

(a) x +- y? + 2° — R? = 0, and (258) is, 
(b) ax + by + cz —R?= 0, 

Eliminate first z and then a between (a) and (6), and (260), (261) 
are ellipses. 

Ex, 2.—Determine the locus of such a point of tangency in sur- 
faces (M), Proposition LI. 


PROPOSITION LIV. 


A plane is drawn through two given points and touches 
. given surface, determine the point of tangency. 


ea 


Let (a,b,c) and (d,e,f), be the two given points. Since the 


LOCUS OF THE CENTRE OF A SPHERE, 117 


tangent plane passes through (d,e,f), its equation for that point is, 

(262) f—z=p(d—z)+p(c—y), 

and we have, (258), (259), and (262), three equations to find z, y, 

and z, the co-ordinates of the point of tangency required. 
Ex.—Determine such a point of tangency in surfaces (M), Pro- 


position LI. 


PROPOSITION LV. 


A sphere passes through a given point, and touches a 
given surface, determine the locus of its centre. 

Let (a,b,c) be the variable co-ordinates of the centre of the 
sphere, then the equation of the sphere is, 

(263) (a a)? + (y —b) + (2 —c) = RB 

Let (d,e, f) be the given point through which the sphere passes. 
For this point the equation of the sphere (263) becomes, 

(264) (d — a)? + (e— b)? + Cf —~ cya" R*. 

Let the equation of the given surface be represented by 
(265) o(,Y,2,) = 0. 

Let p and q, represent the partial differential coefficients of (263) 
and p’ and q’, the partial differential coefficients of (265), 

At the point where (263) touches (265), if we suppose a plane 
tangent to (263) drawn, it will be tangent to (265); and since the 
partial differential coefficients of (263) and (265), at the point of 
tangency, express the tangents of the angles which the traces of the 
same tangent plane make with the axes of x and y, we have, 

(266) p = Pp; and 
(267) Gi ai 

At the point of tangency the co-ordinates 2, y and z, are common 
to the four equations (263), (265), (266), (267), eliminate these 
three co-ordinates between these equations, Represent the resulting 
equation by 
(268) ¢(a,6,c, R) = 9, 
and we have, (268) and (264) to eliminate the variable radius R. 


The result of this elimination, is an equation containing constants 
1 


118 DIFFERENTIAL CALCULUS. 


and the co-ordinates (a,b,c) of the centre of the sphere. Repre- 
sent this equation by 

(269) 1(a,6,c) = 0, 

and we have the equation of the locus required. 

We might with more brevity say that (269) is obtained by elimin- 
ating the four quantities a,y,z, and R, between the five equations 
(263) (264), (265), (266), and (267). 

Ex. 1.—Let the given surface be the sphere whose origin is at 
the centre, find the locus of the centre of (263), 

Here (265) becomes, 

(a) a ty? 4+ 2 = 3, 

equating the partial differential cofficients of (a) and (263), we have 
for (266) and (267), the equations, 

(5) cme 02, and cy = bz, 

Eliminate 2,y,z, and R, between the five equations (2638), (a), 
(b), and (264), and we have, for the locus required, 

2 ee 2 g (ad, 3) a (i ae 
(c) a? (1 z)+ ee: d?) eer 

When d is zero (c) is a sphere, when d is less than s, (c) is an 
ellipsoid of revolution, when d is greater than s, (c) is a hyperboloid 
of revolution, and when d equals s, (c) is a point. 

Ex. 2,—Let the surface (265) be a plane passing through the 
origin, find the locus of the centre of (263) passing through a given 
point and touching this plane. 


PROPOSITION LVI. 


A sphere touches two given surfaces, determine the locus 


of its centre. 
Let (a,b,c) be the co-ordinates of the centre. Let 


(269) a(x y' 32 ) = 0, 
be one surface, and 
(270) (x'',y'',z") = 0, the other. 


Let the equation of the sphere for the point where it touches 
(269), be 


LOCUS OF THE CURVE OF A SPHERE. 119 


271) (x — a)? + (y' — bf + (7 —c)? = R’, 
and for the point where it touches (270), 
(272) (x"’ — a)? + (y” — bP + (2 —ceP = R*. 


Let p and q be the differential coefficients of (271) 


P and Q Do. of (269), 
p and q’ Do. of (272), 
P’ and Q' Do. of (270). 
Then as in the previous proposition, at (266), and (267), we have 


the equations, 
(273) pe P™ and = @} 
(274) ee a ad Gq” eee 

In (273), p,g,P and Q are functions of 2',y’,z', and in (274), 
pq, P’ and Q’, are functions of 2’,y"', and z.”' 

From the four equations (269), (271), and (278), eliminate the 
three co-ordinates 2',y', and z', The resulting equation will con- 
tain a,b,c, and R, and may be represented by 
(275) F (a,6,o;5R) = o. 

From the four equations (270), (272), and (274), eliminate 2", 
y ,2'. The resulting equation may be represented by 
(276) f (a;6,e,R) = 0: 

Between (275) and (276), eliminate the variable radius R. Tbe 
resulting equation may be represented by 
(277 ¢(a,b,c) = 0, 
which is the locus required. 

Ex.—As an example, let (270) be a plane coinciding with the 
piane of ZY, and (269) a sphere whose equation is, 

(a) (x — my + y? + 22? = m, 

its centre being on the axis of 2, and its surface passing through the 
origin. We have at once for (276), 

(5) Ghee. 

and equating the partial differential coefficients of (a) and (271) 
we have, 

(c) cy = bz, c(x —m) = (a— Mm) 2%. 

Eliminate «,'y,'z', andR, between the five equations (271), (a), 
(c), and (6), and we have, 

(d) b? -+ c? = 4ma, 


120 DIFFERENTIAL CALCULUS. 


Which shows that the locus is the surface of a paraboloid of re- 
volution passing through the origin. 


PROPOSITION LVIL. 
Determine the differential of an implicit equation. 


We have already defined an implicit equation to be one that is not 
solved for any particular variable, and we have deduced the differ- 
ential coefficient from implicit equations, where they were of a 
simple form. The only object of this propoposition is, to determine 
a convenient process for differentiating and obtaining the differentia] 
coeficients when the equation is complicated. 

Suppose we have the equations of curves, 


y — px = 0 (a) 
(N) e+ y—r? =o (b 
ay? + bx? — a2 b? = o fy 


We observed (in Proposition II.) that all such equations may be 
represented by the form, 


(278) O(r,y) = 0. 
We may also represent equations (N) by the still simpler form, 
(279) “sO, 


in which w is considered as containing « and y, or as put for all such 
equations as (N). 

If we differentiate the second of (N) or () for y constant, we have, 
(d) RIG ==" 0. 
which may be called the partial differential of (b). 

In like manner we may take the partial differential of (279), and 
write it, 

dudx 

(e) a ie: ? 
where by introducing dx into the numerator and denominator, we 
signify that (279) has been differentiated for x variable. 

Again differentiate (b) for x constant, and we have, 
(f) 2y dy = 0, 
which is the partial differential of (2) in respect of y, 
In like manner the partial differential of (279) is, 


IMPLICIT EQUATIONS. 121 


(2) dudy __ 2 
dy 

where dy is introduced in the same manner, and for the same pur- 

pose as dz in (e). 
If we add (d) and (f), we have, 

(h) 27 dx + 2y dy = 0, 

which is the total differential of (0). In like manner add (e) and 

(g), and we have for the sum of the partial differentials, 

(280) du dx iv du dy iy 
dx dy 

which is the total differential of (279). 
Divide (280) by dz, and we have, 

du du dy ee, 


281 Jame te oes ; 

Si dx dy dz 

from which we get 

(282) lips “age Se dla 
dx dx dy 


This is the general form of the differential coefficient of (279). 

Since we have assumed u = 0, as the representative of (278), 
we may, in using the form (282), for obtaining the differential co- 
efficient, assume w equal to the equation of the curve. 

Thus in the case of the circle, put w equal to the second of (N), 
that is, 
(k) us=erty—r=o, 

Differentiate (#), first for y constant, and we have, after dividing 
by dz, 


lu 
(1 Ge ass Woe 
4) ae 
Differentiate (k) next for x constant, and we have, 
(m) du = 2Y- 
dy 
Substitute (7) and (m) into (282), and we have, 
(7) dy ae a 
dx y 


which is obviously the same that would be deduced from (A), by the 
usual rules of Algebra. 


122 DIFFERENTIAL CALCULUS. 


As an application of the form (282), let us take Proposition III., 
which proposed to determine the equation of a tangent line to a 
plane curve. 

In (65), we found the equation of the tangent line to be, 

(282a) y —y = p (#' —2). 

For p in this, put its value in (282), and we have for the equation 

of the tangent line to a plane curve, 


du du 
2826 Dee Tap ose — (x — z) = 0. 
(2825) FF (y y) + 7 (x ty =u 


Put into this the values of 2% and pte deduced from the equation 
dy da 
of the curve, u = 0, and we have the equation of the tangent line 
required. 
Ex. 1.—Determine the equation of the tangent line to a parabola. 
Put w equal to the equation of the parabola, and we have, 


(2c) “uy? —Amzr =o. 

Differentiate this for 2 constant, and we have, 
du 

(2d sg —— ehh | 5 

(2d) dy ae 
Differentiate (2c) for y constant, and we have, 

du 

2€ Ee, ag 

(2e) oF 


Put into (2825) the values of the differential coefficients at (2d) 
and (2e), and we have for the equation of the tangent line to the 
parabola, 


(2) 2y (y' — y) — 4m (2’ — 2), 
or adding the double of (2c) to (2f), we have, 
(2g) yy — 2m (v' + 27) = 0. 


Determine in this manner the equation of the tangent line to the 
curves, 
ay + B2’—a@ b= 0, ry —y + ay — bx = 0, 
Let us next take an implicit equation of three variables. Every 
such equation is the equation of a surface, and may be represented 
by the form, 
(283) o(2,Y,%) = 0. 


IMPLICIT EQUATIONS. 1238 


We may also represent any surface by the still simpler form, 

(284) “os 0, 

in which u is supposed to contain x, y, and z, and is a representative 
of (283), or of any equation of three variables. As (284) contains 
three variables, it will have three partial differentials. Using the 
same notation as in (e), and (g), we have for the form of the 
differential of (284), 

(285) du dx °s du dy du dz 


=~) O, 


dx dy dz 
the sum of the three partial differentials. 

If we suppose y constant, in (284), (which makes the second 
term of (285) zero), we have for the partial differential of (284), 
divided by dz, 

(2852) CLE at hig hs 
dx dz dz 

By making x constant in (284), (which makes the first term of 

(285) zero), we have, for the partial differential of (284), divided 


by dy, 


(2856) Civecxy Brae FM og) 
dy dz dy 
From (285a) we have, 
/ du 
ed dx 
(286) ge = ES 
dz 
and from (2856) we have, 
du 
deo data 8 dy 
(287) i = Saabscs q: 
dz 


To obtain the partial differential coefficients (286), (287), we may 
assume uw equal the equation of the surface, and deduce from this 
equation the several differential coefficients that enter (286), (287). 

As an example, let the surface be the sphere. Then putting x 
equal to the equation of the sphere, we have, 


—%2 


’ 
win | 
1 
if 
A 
e it 
V 


s Bry, 
ass 


3 + 
Se 


124 DIFFERENTIAL CALCULUS. 


(r) u=vrtyt 2—p7? — 9g, 
and differentiating this first for 2 variable, next for y variable, and 
finally for z variable, we have, 


(s) a me He oh si = 2y, a a 1 
dx dy dz 
Substitute the several values of (s) into (286), (287), and we haye, 
(t) Seb ei. and i 
dx z dy r 


which are the partial differential coefficients of (7). 

As an example of the application of the forms (286) and (287), 
let it be proposed to determine the equation of a tangent plane to a 
curve surface, 

If the differential coefficients (286), (287), be substituted into the 
equation of the tangent plane (229), that equation becomes, 

th du mie thea Oey er" dus eee 
288) EW At Zw + Be —ano, 


which is the equation of the tangent plane to any surface (284). 
To apply (288) to find the tangent plane to the sphere, substitute 
into (288) the values of the differential coefficients (s), and we have, 
after adding the double of (r) to the result, 
(v) va’ + yy’ + 2z' —r? = 0, 
(as in Proposition XLV., (c), ). 
In like manner, find the equation of the tangent plane to the sur- 
faces, 
e+ y — m2 = 0, 
(P) m* (y — bz)? + n® (2 — az)! — mn? = 0, 
(t — a)’ + (y— b)? + (27 —cP —R2? = 0, 
of which the first is the cone, the second the oblique elliptic cylinder, 
and the last the sphere. 


PROPOSITION LVIII. 
Determine the angles which the normal line at a given 
point on a curve surface makes with the co-ordinate axes. 
If the equations of the projections of a line in space be 


ANGLES OF THE NORMAL WITH THE AXES, 125 


(289) y= be +m, andx=az+n, 

and 6,9’, and 9"’ be the angles which this line makes with the co- 
ordinate axes X, Y,and Z, respectively, it is shown in all books on 
Analytical Geometry, that, 


(290) cos. = ogee 72. CcOs.o = —. “ 1, 
(a? + 6° + 1)* (a? + 6 + 1)° 


1 
@ +e Fi" 

If the line in space be the normal, the equations of its projections 
are by (239), 

(291) 2 —x=—p(z'—z), andy’ —y = q(x’ —2z). 

In (291), —p and —q take the place of a and b in (289). Hence 
for a and 6 in (290) take the values —p and —gq, and we have the 
angles requized in terms of the differential coefficients of the given 
point on the surface. If for p and q we take their values in (286), 
(287), the angles (290) become, 


and cos.9@'’ = 


du du du 
(292) COS.6 = 0 —__, COS.0 = b —, 'cOs.0" = B —, 
dx dy z 


where we put for ee 


(293) = }(Z)+ G) + + (BYE 


Ex. it stop tae the surface be a sphere, find the angles the 
normal to the sphere makes with the axes. 
Substitute the values (s) last Proposition into (293), (292), and 


putting the value of v from (293) into (292), we have for the angles: 


required, 
x 2 % 
(w) co.9-= ~, cose’ = 27, cosp’ =~. 
* | r r 
Ex. 2.—Find these angles in the surfaces 
a? +. y? —4mz = 0, ety —m2=0o 


K 13 


iS = eat Bre 
eS 


or te 


Pet ms 


at 


= 
33 


DIFFERENTIAL CALCULUS. 


CONSECUTIVE SURFACES. 


Tue nine following Propositions will be devoted to the discussion 
of Consecutive Surfaces ; and though each Proposition may of itself 
fail of that generality which is possessed by those hitherto discus- 
sed, yet taken together, they form an interesting part of the Geome- 
try of Surfaces, and exhibit much of the power of the Differential 


Calculus. 


PROPOSITION LIX. 


The centre of a sphere moves along a given line, deter- 
mine the locus of the intersection of the sphere with its con- 
secutive sphere. 

We will illustrate this by taking a few particular cases, in which 
it will be seen that the line on which the centre moves, determines 


the nature of the locus of the intersections. 


Ex, A. 

Suppose the line on which the 
wentre moves be the axis of X. 

If N be the centre of the 
sphere in one position, and 
AN = a, then 
(294) (oa ty + 2 = RY 
is the equation of the sphere. For the equation of the consecutive 
sphere, increase AN by NS = da, and we have from (294), 

(295) (x — a— day+y+2= R’ 

Where the two spheres (294), (295), intersect, the co-ordinates 
x,y, and z, are common. If we subtract (294) from (295), the co- 
ordinates x,y, and =, of the resulting equation will appertain to the 
line of intersection of the two spheres. But subtracting (294) from 
(295), and making da indefinitely small or zero, is obviously differ- 
entiating (294) for a variable. Differentiate, therefore, (294), for a 
variable, and we have, 


CONSECUTIVE SURFACES. 127 


(296) — 2(n'— a) da = o. 

In (296) the co-ordinates of the surface, (in this case 2) appertain 
to the intersection of (294) with its consecutive surface. If now 
we eliminate a between (294) and (296), the resulting equation 
which is, 

(297) y+ 2 =" R?, 
is the equation of the surface, which is the locus of the intersection 
of the sphere in any position with its consecutive sphere, Equation 
(297) is the equation of a cylinder perpendicular to the plane ZY, 
and whose axis coincides with the axis of X. 

Ex. B. 

Suppose the line, on which the centre moves, pass through the 
origin, and lie in the plane XY. Let its equation be, 
(298) A. = ma. 

When the centre of the sphere is at any point (a,8) of the line 
(298), its equation is 


(299) (ec —a)+ (y—sy + 2 = R* 
By virtue of (298), (299) becomes, 
(300) (2 — a)’ + (y— ma)? + 2 = R% 


In this, the parameter a determines the position of the sphere. 
Differentiate (300) for a variable, and we have, 
(301) — 2 (x — a) da — 2m (y — ma) da = o, 

The co-ordinates of the surface in (300) and (301) are common 
on the line of intersection of (300) with its consecutive sphere. 

Eliminate the parameter a between (300) and (301), and we have, 
(302) (y — ma)? + 2 = R* 

This is the equation of the surface, which is the locus of the in- 
tersection of the sphere, in any position, with its consecutive sphere. 

If m = 0, the line (298) coincides with the axis of X, and (302) 
becomes (297). 

Ex, C. 

If instead of a straight line, (298), the centre of the sphere moves 
along any curve in the plane XY, then representing that curve by 
(303) 6B = 9a, 
the equation of the sphere (300) becomes, 

(304) (v7 — ay + (y— gay? + 2 = R* 


128 DIFFERENTIAL CALCULUS. 


Eliminate a between (304) and its differential, for a variable. ‘The 
result is the equation of the surface, which is the locus of the inter- 
sections of (304) with its consecutive surface. 

If (303) be a circle whose radius is r, and centre at the origin, 
the surface which is the locus of the intersection of the sphere, in 
any position with its consecutive sphere, Is 
(305) }r + (a? + 9?) - 4 22 = RY 

The line of intersection of a surface with its consecutive surface, 
is called, «« The Characteristic,” a name proposed by M. Monge. 

Corrol. 1st—The characteristic, when the surface is a sphere, (as 
in the previous examples), is a great circle of the sphere whose plane 
is perpendicular to the line on which the centre of the sphere moves. 

For let the line be in the plane of XY, and let (303) be its equa- 
tion, and (299) the equation of the sphere in any position. Differ- 
entiate (299) for a and £4 variable, and we have, 


da 
(306) y—B Fe (x — a) 


The co-ordinates 2 and y, of this equation, are co-ordinates of the 
characteristic. Since z is absent from it, (306) may be regarded 
as the projection of the characteristic on the plane of XY. But 
(306) is the equation of a straight line normal to the curve (303), 
[see (67) ], at the point (a,¢). But the point (a,8) is the centre of 
the sphere (299). Hence the plane of the characteristic passes 
through the centre of the sphere, and is perpendicular to the line 
(303), on which the centre moves. The characteristic is therefore 
uw great circle of the sphere. 

From this it is evident, that the surface which is the locus of 
these characteristics, is tangent to the sphere in every position of it. 

For let 
(307) o(@,y,%) = 0, 
represent the surface obtained by eliminating a between (304) and 
its differential, for a variable, (307) touches the sphere in all its po- 
sitions, the line of contact evidently being the characteristic. Asa 
particular case of this, (279) is a cylinder circumscribing (294). 

In like manner, (305) is the surface tangent to the sphere in every 
position. 


CONSECUTIVE SURFACES. 129 


Corrol. 2d.—Let, in general, 
(308) wit ‘0; 
be the equation of a surface containing the variable parameter 4, and 
the cordinates x, y, and z. If we differentiate (308) for a, y, and 
2 constant, and 6 variable, the differential, which being divided by 
dg, may be represented by 


(309) — = 0, 


will contain the co-ordinates 2, y, and z, which are common to (308), 
and (309), on the characteristic, or line of intersection of (808) with 
its consecutive surface. If ¢ be eliminated between (308), (309), 
the resulting equation may be represented by 
(310) ms 0. 

This equation contains z,y,z, and is the locus of the characteris- 
tics of (308), 

Corrol. 3d.—The locus of the characteristics of (308), viz. (310), 
is tangent to (308), 

For differentiate (308) for 8 constant, and we have for the partial 
differential coefficients, [as in (285a), (285d) J, 

du du du du 

(311) Fo tgs Pia and Gh age OS 

Again, differentiate (308) for x, y, z, and @ variable, and we have 
for the partial differential coefficients, 
(312) soi PO hak dg = 0, and aha du fi jle8 

da? dg dz dy dz dp dy dy 

The a, y, ‘a 2 of (811) and (312), are common on any charac- 
teristic, or on the locus (310), But for any characteristic, (309) 
exists, and (309) renders (311) and (812) identical. Hence p and 
q being the same in (311) and (812), the tangent plane to (308), at 
any point common to (3808) and (310), is a tangent plane to (310). 
The surfaces (308,) (310), are therefore tangent to each other. 

The surface (310) is generally called, The Envelope of the sur- 
face (308), 

Corrollary 2d shows the mode, in general, of obtaining the envel- 
ope of a surface, and Corollary 3d proves the general proposition, 

i3* 


130 DIFFERENTIAL CALCULUS. 


that the envelope is tangent to the surface whose consecutive inter- 
sections, or characteristics, form the envelope. 

As a further elucidation of these principles, take the following 
Proposition. 


PROPOSITION LX. 


The centre of a sphere moves along a given line, its radius 
varies as a given function of the distance of the centre from 
a given point, determine the surface which touches and en- 
velopes the sphere in every position. 

Instead of writing down the general form of the solution to this 
Proposition, we will particularise one or two examples. 


Ex, As 

Let the line on which the centre of the sphere moves be the axis 
of X, and let the radius vary as the distance of the centre from the 
origin. 

Let a be the distance of the centre of the sphere in any position 
from the origin, and since the radius varies as this distance, we may 
put R = ma, where m is any given constant, and the equation of the 
sphere is, 

(313) (x —ayP + y? + 2? = ma’. 

Differentiate this for @ variable. 

(314) -- —(x—a) = ma. 

Eliminate a between (313) and (314), and we have, 


29772 
(315) yt eae 


1—mni ’ 
‘vhich is the surface required. Equation (315) is the equation of a 
cone. 
Ey .B. 

Let the centre of the sphere move on the axis of X, and its radius 
vary as the square root of the distance from the centre to the origin. 

Here put 
Rte (ma)*s 


CONSECUTIVE SURFACES. 131 


which expresses the given law of variation of the radius, and the 
equation of the sphere in any position is, 
(316) (x —ayvr+y + 2 = ma. 

Eliminate a between (316) and its differential for a variable, and 
we have, 


317 gy? tg? = me + cation 
( y” 4 


the enveloping surface, which is a paraboloid of revolution. 
Ex, C, 

Let the line on which the centre of the sphere moves be any curve 
in the plane of XY, and let the radius vary as any function of the 
abscissa of the centre. 

Let (a,8) be the co-ordinates of the centre, and 
(318) B = ga, 
be the curve on which the centre moves. 

Let R = fa denote the function which the radius is of the ab- 
scissa of the centre. Then the equation of the sphere in any posi- 
tion is, 

(319) (v—ay’+ (y—py + # = (fa). 

Substitute into (319) the value of 4 in (318), and we have for the 
equation of the sphere, involving the single Parana: a, 

(320) (x — a) + (y — qa) + # = (fay 

For any particular values of fa and 9a, we can eliminate a be- 
tween (320) and its differential for a variable. ‘The result is the 
envelope of (319). 

Boee. 

If the radius in the preceding example varied as a given function 
of the co-ordinates of the centre, then denoting this law of variation 
by R = 4(a,8), we have for the ‘iti of ‘he sphere, 


(321) (e — ape (ys eS = (xan) 

Eliminate @ from (321) by the relation (318), proceed as directed 
in regard to (320), and we get the envelope required. 

By differentiating (3821) (which includes all the previous exam- 
ples), for a and @ variable, we could show as in the previous propo- 
sition, that the characteristic is a great circle of the sphere whose 


132 DIFFERENTIAL CALCULUS. 


plane is perpendicular to the line on which the centre of the sphere 
moves. ‘This fact would show, without the proof in Corrol. 34d, last 
Proposition, that the locus of these characteristics is a surface touch- 
ing and enveloping the sphere in every position. 


PROPOSITION LXI. 


Through a given point a plane‘ is drawn, cutting off on 
the plane of XY a triangle of given area, determine the 
surface to which the plane is always tangent. 


FIG. 59. 
For simplicity, let D be 4 


the given point on the axis 
of Z, and let 
(322) z= —ma—ny-+p, 
be the plane in one posi- 
tion CDB, cutting off the 
given triangle ABC. Put 
a’ = area ABC, which is 
given. 

For x and y zero in 
(322), we have, 
f = p= AD. For x and z zero in the same, we have, 


D 


y = 2 = AC. For y and % zero, we have, 
(A) n 
« = 2 — AB. For the area of the triangle ABC, we 
m have, 
2 
(323) as! . 
2mn 
From which we have, 
m= —?, 
2a°n 


Substitute this value into (822), and we have, 
324 cs) OE 5, : 
(324) aa Me oie § 


CONSECUTIVE SURFACES, 133 


This is the equation of the plane involving the arbitrary para- 
meter n. Differentiate (324) for m variable, and we have, 


(325) o= pstin — ydn, 


2 


2a7n? 

The co-ordinates a and y in (325) appertain to the charac- 
teristic or line of intersection of (324) with its consecutive plane. 
Eliminate n between (324) and (325). The result is, 

(326) a’? (z —p)?* — p* zy = 0. 

This is the surface which is the locus of all the characteristics of 
(324) and according to Corrol. 3d, Proposition LIX, touches the 
plane (324) in every position. 

It is evident that the characteristic will in this case be a straight 
line, whose projection on the plane XY is, [from (325) ], 

(327) pa — 2a’ny = 0. 

Equation (327) designates a straight line through the origin. 
Hence, all the characteristics intersect the axis of Z, and pass 
through D, 

The obvious properties of surface (326) are, that all sections 
parallel to ZX or ZY are parabolas, and all sections parallel to the 
plane XY, are hyperbolas, whose asymptotes are the intersections 
of the plane of the section with the planes ZX and ZY. 

The nature of this surface might have been inferred from (187). 


PROPOSITION LXII. 


Through a given point on the axis of Z,a plane is drawn, 
whose traces make with the co-ordinate axes, triangles on 
the planes ZX and ZY, the sum of whose areas is given. 
Determine the surface to which this plane is tangent. 

Let the given point be at D, on the axis of Z, and let (322) be the 
plane CDB, in one position. 

Let s°? = the sum of the areas DAB + CAD, Using the lines, 
(A), last Proposition, we have for the sum of these areas, 


2 2 
(328) : = BP. + P * 
2m 2n 


134 DIFFERENTIAL CALCULUS. 


By means of (328), eliminate m from the plane (822), and pro- 
ceed as was done with (324), The result is, 
(329) 239(% —p) = —p (a -|- y*). 

This is the surface required. Its other properties are readily de- 
duced from its equation. 


PROPOSITION LXIII. 


A plane is drawn, cutting off, with the co-ordinate planes, 
a given pyramid. Determine the surface to which the plane 
is always tangent. 


FIG, 60 


Let DCB be the plane in 
one position, and (322) 
its equation. 

Let s* = the solidity of 
the pyramid formed by 
plane (322)avith the co- 
ordinate planes. Using 
the values (A), ineProposi- 
tion LXI., we have for the 
volume of the pyramid, 


(330) 


Substitute the value of m from (330) into (322), and we have for 
the equation of the plane DBC, which cuts off the given pyramid, 
(331) GLE (ae, Sa api arn 9 

6s°n 

As the plane is not limited to pass through a given point, the par- 
ameters 7 and p are both variable in (331), 

It is evident that m and p are independent variables; for n may 
vary, giving a plane which fulfils the conditions of the Proposition, 
while p remains constant; and p may vary, giving a plane, which 
fulfils the same conditions, while m remains constant. 

Differentiate, therefore, (331) for n variable, and we have, [after 
freeing from fractions], 


CONSECUTIVE SURFACES. 135 


(332) 0 = pxrdn — 68°n’y dn. 
Again, differentiate (331) for p variable, and we have, [after free- 
ing from fractions], 


(333) 0 = — xp’ dp + 2s8’n dp. i 
The co-ordinates a, y, and z, in (331) and (832), are common on ; 


the characteristic, or line of intersection of (331), with its consecu- 
live plane made by n variable. 

The co-ordinates x, y, and z, in (331) and (333), are common on 
the characteristic, or line of intersection of (331), with its consecu- 
tive plane made by p variable. 

Hence at the point of intersection of these two characteristics, the , 
co-ordinates x,y,z, are common to (331), (332), and (333). f | 

If, therefore, n and p be eliminated between these three equations, | 
the co-ordinates of the resulting equation will appertain to the point | 
of intersection of three of the planes consecutive two and two, The a 
locus of this point is therefore the surface required. 

Its equation, found by eliminating m and p between (331), (332), 

(333), is 
(334) Cyst et $*. 
9 

The obvious and remarkable properties of this surface are, 

That it 1s asymptotical to each of the co-ordinate planes: That 
every section parallel to any of the co-ordinate planes is a hyper- 
bola, whose asymptotes are the traces of the plane of the section, 
on the co-ordinate planes. 

By a similar process, would be solved the two following problems. 

A. 

A plane is drawn cutting the co-ordinate planes so that the sum 
of the areas of the triangles formed by its traces with the co-ordi- | 
nate axes is a constant quantity. Determine the surface to which 
the cutting plane is tangent. 

B. 

A plane is drawn, cutting the co-ordinate planes so as to form with | 
its traces a given triangle CBD. Determine the surface to which rh | 
the plane is tangent. el | 
The algebraic detail of these Propositions presents some difficulty 


136 DIFFERENTIAL CALCULUS. 


PROPOSITION LXIV. 


. Determine the relation between the parameters of a plane 
which passes through a given point, and touches a given 
surface. 

Let the given point be at the origin, Let the equation of the 
plane be, 


(335) % = mx + ny, 
and the equation of the surface, 
(336) o(2,4,%) = 0. 


It is required to determine the relation between the parameters m 
and n, when (3385) touches (336). 

At the point of tangency, the co-ordinates are common in (335) 
and (336). If p and q be the partial differential coefficients of 
(335), and P and Q of (886), at the point of tangency, we have, 
evidently, 

(337) Pash Sans cant), 

At the point of tangency, the co-ordinates x, y, and z are com- 
mon in the four equations, (335), (336), (337). Eliminate these 
co-ordinates between these equations. The resulting equation will 
contain the parameters mand n. Solving it for m, we have, 

(338) m = on, 
which is the relation required. 
Cor.—If we substitute (838) into (335), we have, 
(339) % = one + ny, 
which 1s the equation of a plane tangent to surface (336), involving 
only one parameter, 7. 

If we eliminate the parameter n from (339), by the principles of 
consecutive surfaces, the resulting equation, which may be repre- 
sented by 
(346) (237) 2) ="; 
will be the envelope, or locus of the characteristics of (389), 

If we define a Developable Surface to be one, whose tangent 
plane, at any point, coincides with the surface along a straight line, 
(340) is a developable surface. 


CONSECUTIVE SURFACES. 137 


As a particular example, let the surface (336) be a sphere whose 
equation 1s, 
(a) (x —a)’? + yet 2 = R* 

The relation between the parameters m and n, of the tangent 
plane (335), is found, by the foregoing process, to be 


(b) oo (se): 
C= is 


Substitute this value of m into (335), and we have the tangent 
plane involving only one parameter 2.. Eliminate the parameter 7 
from this tangent plane, by the principle of Consecutive Surfaces, 


and we have, : 
a fe mp2 
(¢) ee a 
which is the developable surface formed by the characteristics of 
(335). 

Equation (c) shows this developable surface to be a cone. 

If instead of taking the origin to be the point through which the 
tangent plane (335) passes, we were to take any other point whose 
co-ordinates are (a,b,c), then instead of (335), we would have, 
(340a) s>—c=m(x—a)+n(y — 4), 
for the tangent plane, and the solution would be as before. 


PROPOSITION LXV. 

Determine the relation between the parameters of a plane 
tangent to two given surfaces. 

Let the equation of one surface be, 
(341) o(2',y',2') = 
and the equation of the other, 
(342) J(x'',y'',%'') = 0 

Let the equation of the plane touching these two surfaces be, 
(343) 2= me + ny + B. 

It is required to determine the relation between the parameters 7 
n and B, when (343) touches (841) and (342). 

At the point where (343) touches (341), the co-ordinates are com 
mon to both equations, and (343) may be written, 


t 


(344) i me + ny’ + B. 


138 DIFFERENTIAL CALCULUS. 


At the point where (343) touches (342), (843) may be written, 


(345) z'' = ma’ + ny" + B. 
If p and q be the partial differential coefficients of (344), 
If p' and q’ és i 3 (345), 
If Pand Q sa fc a (341), 
[f P’ and Q’ oc ss is (342), 

then we have, as at (266), 

(346) p= RB; q = Q, 

and also, 

(347) p=P,d =Q. 


In (346) the partial differential coefficients are functions of 2’ y’ 

and z', and in (347) they are functions of x" y" and 2", 
Eliminate 2',y',2' from the four equations (341), (344), (346). 

The resulting equation will contain m,n, and B, and may be re- 
presented by 
(348) F (m,n,B) = o. 

Eliminate x" y'' z'' from (842), (845), (847). Represent the re- 
sulting equation by 
(349) f (nm,B) = o. 

Solve (848) and (349) for two of the parameters, as B and x. 
These solutions may be represented by 
(350) B= om, and n= zm, 
which are the relations required. 

Corr. If we substitute from (350) into (3438), the values of B and 
n, we have, 
(351) z= me + nmy + om. 

This is the equation of a plane tangent to the two surfaces (341) 
(342) and involving the parameter m. 

By putting 

d.nm ~.. 


= nm, and fa he =9'm, 
dm dm 
the differential of (351) for m variable, divided by dm, may be written 
(352) rtamy+om=o. 


If m be eliminated between (351) and (352) the resulting equa- 
tion will be the developable surface, which touches the two surfaces 


CONSECUTIVE SURFACES. 139 


(341) and (342), and is the envelope, or locus of the cbaracteristics 


of (351). 
We may represent this resulting equation by 
(353) F (#,9,2%)'= o. 


As a particular example, let the surfaces (341), (342), be the two 
spheres, 
(d) r'? 4 y'? + 2? —RY = 0, 

) Vw a)? + (y" — p+ (2 — of — BR" = 0. 


In this case equations (346) become 


‘ 4 
(€) m= ——, n=—E, 
and equations (347) become 
; 4 a"! bse a y" An B 
QS —_ | es > 
(J) easel z'' — Cc 


Eliminate x’, y', and z’ between (344), (e) and the first of (d), 
and we have, 
(g) R' (m? + n? + 1)? —B=o. 

Eliminate 2'', y’, and z”’ between (345), (f) and the second of 
(d), and we have, 
(h) R" (m? + n? + 1)? — B—ma—ngt+c=o. 

Equations (g) and (%) are what (348) and (3849) become in this 
case. Solve (g) and (4) for B and n, and we have the particular 
values of (850), from which the plane (351) is determined. 


PROPOSITION LXVI. 


Determine the locus of the intersection of consecutive 
characteristics. 

In Proposition LXV, the surface (353) which touches the two 
surfaces (341), (342) may be considered as made up of the inter- 
sections [i. e. the characteristics] of plane (351) with its consecu- 
tive plane. We may regard (351) as the generating surface whose 
consecutive intersections give the characteristics which make up the 
enveloping surface (353), It is obvious that this generating surface 
(351) in each position contains two characteristics, If these two 


a . 2 
Sear -- aS eR oe = 


140 DIFFERENTIAL CALCULUS. 


characteristics intersect, their point of intersection is on three con- 
secutive generating surfaces, 

The co-ordinates in (352) appertain to one characteristic of (351). 
As (352) does not contain x it may be regarded as the projection 
on XY of the characteristic. 

By the theory of consecutive lines if we differentiate (852) for m 
variable, the co-ordinates of the differential will appertain to the 
point of intersection of (352) with its consecutive characteristic. 
[Proposition XXXVII.] Putting, for brevity, 


d.n'm ¥ d.o'm " 
ae ereettery “Tie ATs een me IT 
dm dm 
the differential of (352) for m variable may be written, 
(354) nan’ my + om =o. 


If m be eliminated between (352) and (354), the resulting equa- 
tion will contain x and y, and may be represented by 
(355) F (x,y) = 0. 

This is the equation of the projection on the plane XY of the 
locus of the intersection of consecutive characteristics. Its projec- 
tion on the plane ZX may be found by eliminating m and y from 
the three equations, (351), (3852), (354). Represent the result of 
this elimination by 
(356) F (4%) = 0, 
and we have the two projections (355), (356), which make known 
the locus required. 

This locus is by some mathematicians called the Edge of Re- 
gression of the envelope, because it is the curve on the envelope to 
which all the characteristics are tangent, and consequently is the 
boundary of the envelope. Equations (355), (356), are the Edge 
of Regression of (853). The envelope (353) may, therefore, be 
regarded as made up of the tangents to the Edge of Regression. We 
will employ this consideration in a subsequent proposition. 

It is obvious, that if a particular value be given to m in (351), 
(352), (354), the values of x,y,z determined from these three equa- 
tions will make known a point on the Edge of Regression. The 
curve may in this way be determined by points. 


CONSECUTIVE SURFACES. 141 


PROPOSITION LXVII. 
A plane touches two given curves; determine its en- 
velope. | 
A curve situated in auy manner in space, may be represented by 
its projections on two of the co-ordinate planes. Let XZ and YZ 
be the planes on which the projections are made. Let the projec- 
tions of one curve be, 


(357) Y= 02’, ot eo toh 

and of the other, 

(358) Yom Fx, ee Ee 
Let the equation of the required plane be, 

(359) 2=mx + ny +.B. 


We must first determine two of the parameters m,n,B, in terms 
of the third. For this purpose, observe that where the plane (359) 
touches the curve (357), the co-ordinates are common to the plane 
and curve, and (359) may be written, 


(360) Zz = mz’ + ny’ + B. 
In like manner, where (359) touches (358), we may write (359), 
(361) 2 == mz + ny’ +B, 


Observe, also, that the tangent line to the curve (357), at the 
point of contact of (357) and (359), lies in the plane (359). 

The projections of this tangent line are tangent to the projections 
(357) of the curve, and the equation of the tangent line in space 
will be the equations of the two tangent lines to the two projections 
(357). ‘The equation of these tangent lines are, as at (65), 


(362) y—y= dy »(2—2'), and a—2' — gx » (422°). 
dz dz’ 

These together represent the tangent line to the curve whose pro-. 
jections are (357). 

And since the line (362) coincides with plane (359), we have, 
[from Analytical Geometry], by the condition of the coincidence of 
a line with a plane, 

(368) Z Gri, xe ‘ dy' 


—l—o. 
dz' dz' 


142 DIFFERENTIAL CALCULUS, 


Similarly for the coincidence of the tangent to (358) with (359), 
we have, 
(364) i oe +n dy" —l=o. 
Be) aay: dz" 
Eliminate 2’, y', and z’, from the four equations (357), (860), and 
(363), and the resulting equation will contain m, n, and B, and may 
be represented by 


(365) ¢(m,n,B) = o. 
In like manner, from (358), (361), (364), we have, 
(366) L(mn,B) = o. 


By means of (3865) and (366), two of the parameters may be 
eliminated from (359), and the envelope found, as in Prop. LXV. 
As a particular example, let the curve (357) be, 


(g) y= ks", y — bz'?, 
and the curve (358), 
(h) "= cz", y"" ree ez!2. 


The particular values of (365) and (366), determined as above 
directed are, 
(k) 2B (am + bn) —1 = 0, 2B (cm + en) —1l =o. 
Solve equations (%) for x and B, and their values put into (359), 
we have the form of the tangent plane, in this particular case, in- 
volving the parameter m. ‘This parameter being eliminated by the 
principle of Consecutive Surfaces, we have the envelope required. 


PROPOSITION LXVIII. 


Determine the increment of the ordinate in passing from 
one point to another, on a given curve surface. 


This Proposition leads to the development of functions of twe 
variables, 


DEVELOPMENT OF A FUNCTION OF TWO VARIABLES, 143 


FIG, 62. 


Let the equation of the 
surface PMK be 
(366) x = o(xry). 
For the point L on the 
surface the co-ordinates 
fie AL = 2, ‘TV gy, 
and VL = z., 

In passing from L to P 
on the surface, the or- 
dinate VL is_ increased 
by PR. It is required to 
find the length of PR. 

Through L and P pass 
planes parallel to the co- /z 
ordinate planes ZX and ZY. 

If x receive the increment h = TU, while y remains constant, 
the = in (366) becomes OD, which is the ordinate in the parallel 
section. Put OD = z', and by Taylor’s Theorem, [see (218) ], 
equation (366) becomes 


(367) 2' = ox +hy) = +h ++ te WP+ &e. 
If in (366) y receive an ile k = VB, while 2 remains 


constant then z becomes EB which is the ordinate in the other parallel 


section, Put EB = 2”, and by Taylor’s Theorem, (366) becomes 
lz d’z ; 

368 strate ky =. 2 al) “et k? Wwe. 

(365) o(r,y + k) Ss dy mi Qdy a 


If we transpose z in (367) and (368), we observe, at once the in- 
crement of the ordinate in each of the parallel sections. 

From (367) or (368), we observe, that to pass from one point to 
another on a parallel section, the second ordinate (z' or x'’) is equal 
to the first ordinate, plus the successive differential coefficients of the 
parallel section multiplied by the increment of the abscissa of the 
section. Hence, to pass from O to P, in the parallel section OP, 
put PC = z’”, and by analogy to (367), we have, 


144 DIFFERENTIAL CALCULUS. 


saeiyel d?z' 
Bou dae = 2’ ae 
(369) te i ++, Ty 
To pass from E to P in the parallel section EP, we have in like 
manner, 


dz" d2z"' 
370) zi = gee Ss He OLE, ee h, k). 
(370) Pets oa. + &c. = oe + hy y +h) 


K+ &e. = O(a + hy + k). 


Differentiate (867) for 2 constant and y variable, (y being sup- 
posed to enter into the value of z, d*z, &c.) and we have, 
(370a) Ug DEY gp FP eater 

dy dy dxady 2dax*dy 

Differentiate (870a) again for x constant and y variable, and we 
have [y being the independent variable], 
(370b) oh eg ce 
dy? dy? = dxdy/? 

If we were to differentiate (3706) again, we would have the third 
differential coefficient in respect of y, wc. 


+ & 


For 2", ———9 pee &e., 
dy' dy? 
in (369) put their values in (367), (370a), (8706), and we have, 
d d 
gia o(@.+ hy y+ kya t Sh math ae 
371) i 
BO gpg Oe bh ed 18) wa, 
| 2dx? dx dy dy? 
Were we to differentiate (368) for y constant and x*variable, and 
’ dz" d2z"' 


&c., their values from 


—asiG 2 


——a 9 - 
da" da? 


(368) and its differentials, we have, 


then put into (370) for <’, 


dz sent d’z 


wae vend 
dy 2dz* ‘s 


2'"—e(e+th,y +k) = z+ a ho + 


da hk dz 
| dydx . 2dy? 
The developments (371) and (871a) are equivalent, each being 


the length of the same ordinate PC or 2'”. 
Equation (371) or (371a), is the length of the ordinate PC, in 


(371a) 


DEVELOPMENT OF A FUNCTION OF TWO VARIABLES. 145 


terms of the ordinate at the point L, and of the increments and dif- 


ferentials in the parallel sections through L. 
If we put, 


dz dz d*z d?z d*z 


372 = Ss =). oor 8) G00 = ft, 
ie? de * dy : dx? ” dx dy dy” 
and use only the first six terms of (371) or (371a), we have, 
(373) 2!" —2= ph + gk + rh? + 2shk + tk*)+ &c,, = PR, 


which exhibits the increment of the ordinate in passing from one 
point to any other on a curve surface. 
Ex,—Let the equation (366) be 


(a) 2 = x” y". 
If x receive the increment h, and y the increment k, (a) becomes 
(0) a!" = (x + hy) (y + k)", 


which may be developed by formula (371) or (371a). 
Cor. If the developments (871) and (371a) be equated the co- 
efficients of the like powers of h and k are equal. This gives the 


condition. 
d’z 2a 

373a mee th a aaa 
80) dx dy dy dz 

But the first side of (3710) is the differential of i for y con- 

YY 
stant, and x variable, the second is the differential of as for y va- 
x 

riable and x constant, and (3715) shows these differentials to be 
equal, This may be exhibited in a particular case, as as follows: 


Differentiate (a) above for x variable, and we have, 


a 


1) = nx! a’. 
(¢) a y 
Differentiate (a) for y variable, and we have, 
dz 
(d = ane yf 
(d) 7 y 
Differentiate (c) for y variable and x constant, and we have, 
d*z n—1 »,m—1 
(€) —— Mr yy” > 
dx dy 
Differentiate (d) for x variable, and y constant, and we have, 
ow 
(ft) ar — nm gn} yr, 


dy dx 


146 DIFFERENTIAL CALCULUS. 


Equations (e) and (f) show the values of these differential co- 
efficients to be equal, as in (3716). ‘This principle is of much im- 
portance in the Integral Calculus. 


PROPOSITION LXIX. 


Determine the differential equations of the parallel and 
inclined sections of a curve surface. 


Definitions. 
ist. If the equation of a curve be differentiated the result is called 
the differential equation of the curve. Thus, 2%dx + 2ydy = 0, 


is the differential equation of the circle a? + y*? = R?%, 
FIG. 64. 


2d. We have defined 
(Proposition XLV.) a pa- 
rallel section of a curve 
surface to be one whose 
plane is parallel to one of 
the co-ordinate planes 
ZX or ZY, and a parallel 
trace to be the trace 
on XY, of the plane of the 
parallel section. Let the 
section whose plane is 
perpendicular to the plane 
XY, and oblique to the 
other co-ordinate planes 
be called an Inclined Section. 

Let the trace on XY, made by the plane of the inclined section, 
be called, Lhe Inclined Trace, and let the plane which cuts the in- 
clined section be called, The Inclined Plane. In the figure, if L 
and P be two points on the curve surface, LPCH is the inclined 
plane, CH is the inclined trace, and the section through L and P 
made by this plane, is the inclined section. 

Let the equation of the curve surface be, 


(374) z= 9(#,y). 


SECTIONS OF A CURVE SURFACE, 147 


If (374) be differentiated for x constant, and y variable, we have, 


(375) dz = d.p(x,y).dy = al dy = qdy. 
dy dy 


If (374) be differentiated for y constant, we have, 
d.o(x,y) da i dz dite nat 
dx dx 

Equation (375) is the differential equation of the parallel section 
XL, and (376) of LO. 

If the points O and E be indefinitely near to L, then VD is the 
representative of dx in (376), VB of dy in (3875), QQ of dz 
in (576), and EN of dz in(375). 

For the differential equation of any section through L and P, dif- 
ferentiate (374) for x and y, both variable, and the total differential 
being the sum of the partial differentials, this differential is, 

(377 % = pdx + qdy. 

Here dz would be represented by PR. 

Equation (377) is the differential equation of a section made by a 
plane cutting the surface in any manner, in other words, it is the 
differential of the surface itself, and is called the total differential 
of (374). 

To obtain the differential equation of an inclined section, let the 
equation of the inclined plane PCH be, 

(378) y = Ax + B, 
which is the equation of a plane perpendicular to plane XY. 
From (878), we have, by differentiating 


(379) dy = Adz. 
This value substituted into (377), we have, 
(380) dz = (p + Ag)dz, 


which is the differential equation of the inclined section made by 
plane (378). 
Ex.—Determine the differential equation of the inclined section 
made by plane (378) with the following surfaces : 
x? + y* — Amz = 0, v4 yy —m sz 
(Q) e+t+yt+2—hRh=o, 
a2? + Pa? + FF? 2?7—a’? b? = oo. 
From the first of (Q,) we have, 


2 


= 0, 


148 DIFFERENTIAL CALCULUS. 


dz x dz y 
et — eae ea jo i 0 perch os BP 
dx 2m ae dy 2m 4 
These values substituted into (380), we have, 
dx 
init Ay) ———, 
dz = (x + Ay) + 


for the differential equation of the section. 


PROPOSITION LXX. 


Determine the length of the subtangent at a given point 


on the inclined section of a curve surface. 
FIG. 65. 


Let P be the given 
point, and 
(381) z= 9(*,y) 
the equation of the sur- 
face. Let (378) be the 
inclined plane cutting the 
section, and (380) the dif- 
ferential equation of the 
section. Let PH be the 
tangent line at P to the 
inclined section, then CH 
is the subtangent whose 
length is required, 


If we suppose the in- 
clined trace CH to be the axis of abscissas of the inclined section, 
and call it the axis of S, then CV = dS. If the co-ordinate planes 
be rectangular, CDV is a right-angled triangle, and since VD = dz, 
and DC = dy, we have, 


(382) dS? = dx? + dy’. 
Substitute (879) into (382), and we have, 
(383) dS = dx (1 + A?! 


From similar triangles PRL and PCH, we have, 
(384) PR: RL:: PC: CH, or dz: dS:: z : subtangent. 
By means of (383) and (380), we have, from (384), 


MINIMUM SUBTANGENT. 149 


AN" Bot 
p+qAa 
This is the length of the subtangent required, involving the par- 
tial differential coefficients p and q, which are to be deduced from 
the equation of the surface (381). 


(385) subtang. 


Ex.—Determine the subtangent in the inclined section of sur- 
faces (Q), last Proposition, 


PROPOSITION LXXI. 


Determine the length of the subnormal at a given point on 
ihe inclined section of a curve surface. 


From the point P, fig. 65, draw PG. perpendicular to PH, and in 
the plane of the inclined section PHC. Then CG is the subnormal 
whose length 1s required. 

Since the triangle GPH is right-angled at P, we have, 


(386) GOs 


But PC is the ordinate z. Substitute into (386), from (385), the 
value of CH, and we have, 


(387) subnormal = ae gh) : 
(i + A*)* 
Ex.—Determine the subnormal in the inclined section of sur- 


faces (Q). 


PROPOSITION LXXII. 


Determine the position of the inclined plane through a 
given point on a curve surface, when the subtangent of the 


inclined section 1s a minimum. 
ld 


150 DIFFERENTIAL CALCULUS. 


FIG. 66. 


Putting T’ for CH, the 
length of the subtangent, 
we have from (385), 

2 (1+A?)* 
p+qa- 

For the point P, the 
ordinate z is constant, and 
p and q the differential 
coefficients of the parallel 
sections are also constant. 

Hence for the point P, 
the subtangent (388) is a 
function of A, the tan- 
gent of the angle BCV. 


Differentiate (388) for A variable, and putting its differential equal 
to zero, we find, 
(389) A=4q-=pP. 

This value of A, substituted into (388), we have, 


~ 
“<<, 
. 
~ 


(388) T = 


ee ee 


R 


T = ———_, 
(cher) 

The value of A in (389), gives the position of the inclined plane 
(378), when the subtangent is a minimum, as may be verified by 
substituting, according to Proposition XXXI., (889) into the second 
differential of (388). 

The value of A in (389) changes (378) to 


(390) 


(391) y= tnt B. 
if 
This is the plane of the inclined section whose subtangent CH, is 
a minimum, 
The plane of an inclined section through P perpendicular to 
(391), is + 


(392) ’ eae 


MINIMUM SUBTANGENT. 151 


But the value, 
A=—p-+q 
substituted into (388), renders the subtangent infinite. 

Hence the tangent line at P, to the inclined section made by plane 
(392), is parallel to the plane XY. 

Let us call the inclined section made by plane (391), The Minv- 
mum Section, and plane (391) The Minimum Plane. 

Corrollaries. 

1.—The tangent line at P to the minimum section, is shorter than 
the tangent line to any other inclined section through that point. 

For in the right-angled triangle PCH, if PC remain constant, PH 
is a minimum when CH is. 

2,—The tangent line at P to the minimum section has a greater 
inclination to the plane XY than any other. tangent through that 
point. 

For in the right-angled triangle PCH, if PC remain constant, the 
angle PHC is greatest when CH is least. But the angle PHC is the 
inclination of the tangent line to the plane XY. 

3.—The inclined sections that have the greatest and least tangents 
are perpendicular to each other. 

4,—If a tangent plane to the surface be drawn through any point 
of the minimum section, its trace HF on the plane XY will be per- 
pendicular to the tangent line to the minimum section. For since 
all the tangents to the surface at the point P lie in the same plane, 
and the tangent of the minimum section is the shortest, it is perpen- 
dicular to the trace HF. 

The minimum section is the curve on which a heavy body on a 
curve surface moves when the plane XY is horizontal. This sec- 
tion is often used in Mechanics. 

By substituting into (3880) the value of A in (389), we find for 
the differential equation of the minimum section, 


(393) dz = (p* + 7’) - 


By equating the values of A in (389) and (379), we have, 
(394) pdy+qdr=oa, 


152 DIFFERENTIAL CALCULUS. 


which may also be used as the differential equation of the same 


section. 
Ex.—Determine the length of the minimum subtangent in sur- 


faces (Q). 
PROPOSITION LXXIII. 


Determine the position of the inclined plane through a 
given point on a curve surface, when the subnormal of the 


inclined section is a maximum. 

This is determined by differentiating (387) for A variable, and 
equating the differential to zero. ‘The value of A is found to be, 
(395) A =i Dp, 
which renders (387) a maximum. 

Comparing (395) and (389), we observe, that the inclined section 
which has a maximum subnormal is the one which has a minimum 
subtangent—a result which might have been observed immediately 
from the diagram fig. 66; for in the right angled triangle PGH, if 
PC remain constant, CG is greatest when CH is least. ‘The same 


truth is manifest in (386). 


PROPOSITION LXXIV. 

Determine the second differential of an inclined section 
of a curve surface. 

Equation (377), viz. 
(396) z= pdx + qdy, 
may be taken as the first differential of any section of a curve sur- 
face. It is usually regarded as the differential of the surface, being 
that of a section of the surface made by a plane cutting it in any 
manner. As 2 and y are the independent variables, dx and dy are 
constant, Differentiate (396) for dx and dy constant, and we have, 


(397) .. @z = dp dx + dq dy. 

dz dz 
397a But) ie eee ANG Gia lcd. 
( ) 3 dx : d 


The differential of p is the sum of its partial differentials, first for 


RADIUS OF CURVATURE. 153 


x, and second for y variable. Hence, we have, from the first 
of (397a), 


(398) “ OD = ila dx + eid . dy. 
dx? dx dy 7. 

Using the notation at (372) for the partial second differential co- | | 
efficients (398) becomes F 
(399) dp=rdzx-+s dy. : 

By differentiating g = Ua in the same manner and using nota- i 
tion (372), we have, dy 
(400) dq = sdx + tdy. 

Substitute (399) and (400) into (397), and we have, bd 
401) 2 —rdz + %sdz dy + t dy’. 4 


This is the second differential of any section. 

[t must be carefully observed that 7, s and ¢ in these equations are 
the differential coefficients of the parallel sections, and that d?z on 
the first side of (401,) and (397) is the second differential of z in any 
section, 

By proceeding with (401) as we have here done with (396), we 
should obtain the third differential of any section, &c. 

To obtain the second differential of the inclined section made by 
plane (378), substitute into (401) for dy, its value in (379), and we 
have, 

(402) d’z = r dz? + 2s Adz? + ¢ A®dz?. 

This is the second differential required. 

Ex.—Determine the second differential of the inclined section 
made by plane (378) with the surfaces, | 

x + y?— 4mz = 0, r+ y? — m2" = o. ml 
{ 


PROPOSITION LXXV. “a 
Determine the radius of curvature at a given point on an 
inclined section of a curve surface. 
If 8 be the ordinate, and a the abscissa of any point on a plane 
curve, the radius of curvature is, [see (199), ] 


15* 


SSS 


‘a 
4 


154 DIFFERENTIAL CALCULUS, 


(da? + dpe)? 


403 R = 
ae : da Bary 


If the inclined trace CH so 


“. 
mee 
~ 


be taken as axis of ab- 
scissas, and called the axis 
of S, and the ordinate of 
any point P on the inclined 
section be z, the a and g 
of (403) become S and z 
in the inclined section, 
and (403) becomes, 


AL resem baum 


Substitute into (404) 
for dS, dz, and d?z, their 
values in (383), (380), and (402), and we have, 


iilacd Lihat ay) 


bai 2sA + iA’). (: oe xt 


This is the radius of curvature required. 
If we substitute into (405), for A, its value in (389), we have, 
for the radius of curvature of the minimum section, 


(406) es + P+ a) (p+ 4) | 
rp + 2s pg +i” 
Ex.—Determine the radius of curvature in the minimum sections 
of surfaces, 
x + y?— 4mz = 0, e+ y?—m = 0, 


(405) R= 


The first of these surfaces gives for the radius of curvature of 
the minimum section, 


R=2(m +2)? + m?. 


RADIUS OF CURVATURE. 155 


PROPOSITION LXXVI. 


Determine the radius of curvature of a normal section 
through a given point on a curve surface. 

Definition—A normal section through a point on a curve sur- 
face is one whose plane contains the normal to the surface at the 
point, It is obvious there may be an indefinite number of sections 
through a point containing the same normal. 

This is merely a particular case of the preceding Proposition, and 
the solution may be obtained from (405), as follows, 

Suppose the planes of reference so taken that the normal line to 
the given surface, at the proposed point, be perpendicular to the plane 
XY. ‘Then (378) may be taken as the plane of the normal section. 

Suppose P, [fig. 67], be the proposed point, and suppose the ordi- 
nate PC be normal to the surface, then all the tangent lines to the 
surface at P will be parallel to the plane XY, Consequently for this 
point, we have, 

(407) p= 0, and == 05 

because p and q expresses the tangents of the angles which the 
tangent lines to the parallel sections make with the parallel traces, 
and these angles being zero, or 180°, we have (407). 

The conditions (407) reduce (405) to 
(408) ENE rer ee cn 

r+ 2sA 4- tA? 
This is the radius of curvature required. 


PROPOSITION LXXVII. 


Determine the normal section whose radius of curvature 
is a Maximum. 

The second differential coefficients 7, s, and ¢, appertaining to the 
parallel sections through any point P, [fig. 67], are constant for that 
point, Hence the radius of curvature in (408), is for'the same point 
P, a function of A, Hence by the rule for a maximum, differentiate 
(408) for A variable, and putting the differential equal to zero, we 
have, 


156 DIFFERENTIAL CALCULUS. 


(409) AE A ene, 
§ 


This solved, gives two values of A, one of which gives a maxi- 
mum, and the other a minimum radius of curvature. [See Proposi- 
tion XXXI.] 

By the theory of equations, the product of the roots of a quad- 
ratic equals the last term of the equation. Hence put A’ and A” 
for the roots of (409), and we have, 

(410) A‘A" = —1, 

which exhibits the well known relation of the tangents of perpendi- 
cular lines. Hence the two planes given by the two values of A in 
(409), are perpendicular to each other, and we have the remarkable 
property that, The normal sections of greatest and least curvature 
are perpendicular to each other. 

To obtain the length of the maximum and minimum radii of cur- 
vature, substitute into (408) the values of A obtained by solving (409). 


PROPOSITION LXXVIII. 


Determine the radius of curvature at a given point of an 
oblique section. 


Definition.—An oblique section of a curve surface is one whose 


plane is inclined in any given manner to the co-ordinate planes. 
FIG, 68, 


For convenience, suppose the 
point at which the radius of cur- 
vature is drawn to be at the ori- 
gin, and the plane of XY to be 
tangent to the surface at that point. 
Then the axis of Z will coincide 
with the normal. 

Suppose the plane of the oblique 
section make an angle ¢@ with the g 
axis of Z, and let its trace AS on XY, be the axis of abscissas S, of 
the oblique section, and AZ’, the axis of ordinates of the same section. 


LINE OF CURVATURE. 157 


The radius of curvature in the oblique section is as in (404), 
which may be put in the form, 


ds? dz!” 3 
All R = eee aad J yy. 
Cie d?z' ( a ds? 


: - expresses the tangent of the angle which the 
ds 
tangent line makes with the axis of abscissas, which angle in this 
case, Is Zero, (411) becomes, 
d 9 
(412) Bi eipyiete an, Sst 
d?z;' 
For the radius of curvature R' of the normal section at A, whose 
plane has the same trace AS, we have in like manner, 
2 
(413) Rete 


d?z 


But since 


But if a given distance on the axis of z' be projected orthogonally 
on the axis of z, we have, 
% = 2' COS.Q. 
Differentiating this twice, we have, 
d?z = d?z' cos.o. 
Substitute this value of d?z’ into (412), and we have, 
(414) ies HBS 


2 
Sy 
“a 


Comparing (413) and (414), we observe that the radius of cur- 
vature of the oblique section is the projection of the radius of curva- 
ture of the normal section, on the plane of the oblique section, This 
supposes the trace of the planes of the normal and oblique sections 
on the plane of XY, to be the same. 


COS.9- 


PROPOSITION LXXIX. 


Determine the direction of a line of curvature ata given 
point on a curve surface. 


Definitions. 
ist. Let us call the point on a curve surface at which a normal 


line is drawn, The Normal Point. 
M 


158 DIFFERENTIAL CALCULUS. 


2d. A line of curvature is the locus of the normal points of con- 
secutive intersecting normals. 

This definition is necessary for the reason that consecutive nor- 
mals to a surface do not necessarily intersect. They may be in 
different planes and then no intersection occurs. If they do inter- 
sect, their normal points will be on a curve (viz. the line of curva- 
ture), traced on the surface. 

Suppose P be the given point. The equations of normal line 
through P are, 


(415) 


in which 2’, y' and z' are 
the variable co-ordinates 
of the normal line, and 
x, y, and z, the co-ordin- 
ates of the normal point 
a 

For the consecutive 
normal the a, y, z, p and 
q of (415), change when 
the normal point is chan- 
ged and 2’, y’, and z’, are 
common to (415) and its 
consecutive normal at the 


point of intersection.— 2 
Hence, differentiate (415) for x’, y’ and z’ constant, and we have, 
»Y 

(416) § — dx — pdz + (z' —2) dp =o, 


) — dy — qdz + (2! —z)dq=o. 

By the principle of consecutive lines, 2’, y’, and 2’, are common 
to (416) and (415) at the point of intersection of (415) with its con- 
secutive normal. 

Eliminate x’, y', and z' between the four equations (415), (416) 
and the result will be the relation among the co-ordinates and differ- 
ential coefficients of the normal point when the consecutive normals 
intersect. 


LINE OF CURVATURE. 159 


As x’ and y' do not enter into (416), it is sufficient merely to eli- 
minate z’ from the two equations (416), and we have, 
(417) dp (dy + qdz) — dq (dx + pdz) = o. 

If we substitute into (417) the values of dp and dq in (399) 
and (400), and the value of dz in (396), we will have an equation 


he dy . 
containing z, y and a which may be represented by 


dx 
(418) of y; ) Lp 
dz 
If (418) contain z, let the surface be 
(419) a (x,y), 


and z can be eliminated from (418). 

Equation (418) is the differential equation of the projection of 
the line of curvature on the plane XY. If (418) be divisible into 
two factors, representing these factors by 


(420) F(ew, =) = 0, and f (xy, “) = G, 
dx dx 


these two equations would indicate two lines of curvature on sur- 
face (419) at the same point. 

If we suppose the co-ordinate planes so taken that the normal 
line on the proposed surface may be perpendicular to the plane of 
XY, then p = o, and g = 0, and (417) becomes after substituting 
into it from (899) and (400), 

(421) LOR arity: (By ea 
dx? s dx 


d ; hay oe , 
Here = expresses the tangent of the angle made with the axis 
x 


of X by the tangent line to the projection on XY of the line of cur- 
vature. This is the A of (378). And as in (409), so here (421) 
denotes that there are two tangent lines perpendicular to each other. 
Consequently, at the same normal point there are two lines of cur- 
vature at right angles to each other. 

From the coincidence of (421) and (409), we infer, that the. 
lines of curvature at any point, are tangents respectively to the sec- 
tions of greatest and least curvature at the same point; and that 
the intersections of consecutive normals occur at the centres of 


160 DIFFERENTIAL CALCULUS. 


curvature of the sections of greatest and least curvature. ‘This 
property enables us to determine the absolute length of the maximum 
and minimum radii of curvature at any point on a curve surface 
without changing the position of the co-ordinate axes, as was done 
to obtain (408) and (409). For, if 2’, y’ and 2’ be the co-ordinates 
of the centre of curvature, we have, for the radius of curvature, 
(422) R? = (2' — 2)? + (y — yy? + (2 — zy. 
Eliminate a between the two equations (416), and then by 


means of this resulting equation, and (415), eliminate 2’, y’, and =’ 
from (422). The result will be the maximum and minimum radii 
of curvature. 

As a particular example of the determination of lines of curva- 
ture, take the paraboloid of revolution whose axis of revolution is the 
axis of Z. Its equation is, 


(a) ety? = 2m. 
Derive from (a) the differentials of (417), and we have, 

dz x dx 

b is APR a ee ee d —_— semi 

”) dx m a ‘i m 
dz y dy 

i ay es ae gee Sag 

() dy m z i m 

(d) Gn ee de teeny 

m m 


Substitute these differentials into (417), and we have an equation 
which may be divided into two factors, and which furnishes the 
equations, 


dy _ y 
(°) dx x 

dy _ x 
(f) day 


Equations (e) and (f) are the particular forms of (420). 

Equations (e) and (f) are properly the differentials of the projec- 
tions on XY of the lines of curvature. 

Equation (e) is the differential of a straight line through the ori- 
gin. For every such line is of the form, 


ELIMINATION OF ARBITRARY FUNCTIONS, 161 


(g) y=cr ee perce J. 
dz x 
Equation (f) is obviously the differential of a circle, viz, of 
(h) a a” — R’. 


Hence the projections on XY, of the lines of curvature on a para- 
boloid, are a straight line through the origin, and a circle with its 
centre at the origin. 

This result we ought obviously to have. For the projection on 
XY, of the meridional section AL, of the paraboloid, is a straight 
line AR passing through the origin. And the projection on XY of 
a section of the paraboloid perpendicular to the axis of revolution 
AZ, is a circle having its centre at the origin. 


FIG. 70. 


And we know that normals to the 
meridional section AL, would be nor- 
mals to the surface, and intersect in the 
plane of the meridional section, and that 
normals to the surface on a circle per- 
pendicular to the axis of revolution, 
would intersect on the axis of revolution, 
The meridional section and circle per- 
pendicular to it on the surface are there- 
fore the lines of curvature of the para- 
boloid of revolution. 

The same result would be obtained for any other surface of revo- 
lution except the sphere. For the sphere, all great circles are lines 
of curvature—a result that the analysis above would also make 
known. 


M 


PROPOSITION LXXX, 

Determine a method for eliminating an indeterminate 
function. 

Suppose we have the equations, 
(423) U= f2, and 
(424) 2 = F (x,y), 
in which « and y are independent variables, and fz an indetermi- 

16 


ie 


ue 
» .? 
Be 
"a 
i. 
‘BY 
Vek 
Mid 
, 


162 DIFFERENTIAL CALCULUS. 


nate or unknown function of x. It is required to find a relation be- 
tween u, x, and y, independent of the function fz. 
The differential of (423) is, 


dz 
and the two partial differentials of (424) are, 
(426) dz = Ma dz = o Een 
dx dy 


If for dz, in the numerator of (425), we put in succession the 
values of dz in (426), we have, 
(safiarety ened See cid ios beri se Tal): 

dz dx dy dz dy 
Divide equations (427), and we have, 
du _ du _ dF (ay) . dF (x,y) 

dx dy dx ’ ET ako 
This is the relation required, and is independent of the function fz. 
Ex.—Let (423) be, 


(428) 


(a) u = log(a + 2), 

and let (424) be, 

(0) = mM + ny. 
Then (428) becomes, 

(c) du — du = Bs 


dx = dy n 

This is an equation among the partial differential coefficients, and 
would be the same if, instead of (a), we had any other function of z. 

Again, suppose we have the equations, 

(d) 8 = $a, a =f (2,4), se (Y,2), 
in which z and y are the independent variables. 

If a relation be required among the variables x,y,z, in these equa- 
tions, independently of ga, proceed as follows, [writing df for 
df (x,z), and dF for d F (y,z). ] 

Differentiate equations (d) for y constant, and we have, 


doa.da 
(g) dg= 


ELIMINATION OF ARBITRARY FUNCTIONS. 163 


df.dz df.dx 


h Ld ip Sas Cua 
(*) — dz ¥ dx 
IF dz 
k a. 
(*) ‘ dz 
Again, differentiate equations (d) for # constant, and we have, 

doa. da 
l dg = 
(?) 8 ~ 
(m) da = df.dz 

dz 

(7) dg. = dk Lid dE .dy 

dz dy 


For da in the numerators of (g) and (1), substitute the value in 
(h) and (m) respectively, and we have from (g), 


dpe _—§s_ doa es 1 ¥ 
(°) dx da N\dz da} 
and from (f), 
dg dea df 
(r) dy ip a PRP q- 
From (k) we have, 
dg _ dF 
(8) ee 
and from (7), 
dB dF dF 
t PLB Syste Se nde Ss 
) dy dz a dy 
Divide (0) by (7), and eliminate - and a by (s) and 


(t), and we have, 
dF dF df ,. 4 of 
429 ae tes, ae 
a ae =) ipo ae + ‘ 
This is the i among the partial differential coefficients, in- 


dependently of 9a. 
An application of this general process will be given in the next 


Proposition. 


DIFFERENTIAL CALCULUS. 


PROPOSITION LXXXI. 


Determine the equation of a cylindrical surface. 
Definitions. 

1st.—A cylindrical surface is one generated by a straight line 

moving parallel to itself along a given curve, 
_ 2d.—The curve along which the straight line moves is called, The 

Directrix of the Cylinder. 

Let the curve in space, which is the directrix, be represented by 
the projections, 


(430) Lis OF; Yy = Yr 
And let the equation of the generating line be, 
(431) 3 = mz.+ a, es a= £— mM, 
y= nz + B, Pa oe ates 


nt Since the line is always parallel to itself, m and n are constant, 
al Where the line (431) meets the directrix (480), the co-ordinates 
are common to (430) and (481). Eliminate «, y, and z between 
these four equations, and we have a relation between a and £, which 
may be represented by 
(432) = oa. 

Substitute into this the values of a and 8, in (431), and we have 
for the general equation of cylindrical surfaces, 
(433) y — nz = o(@ — mz). 

If oa be eliminated between (431) and (432) by the principles of 
last Proposition, we have, 
(433a) ng + mp—l1=oa, 
for the relation among the partial differential coefficients of all 
cylindrical surfaces, whatever be the directrix. Equation (433) is the 
differential equation of all cylindrical surfaces. 


PROPOSITION LXXXII. 
Determine the equation of a conical surface. 
Let the equation of the directrix be 
(434) Lio and y = . 


CURVES OF DOUBLE CURVATURE. 165 


And let the equation of the line which generates the cone, be 

(435) ; °— a’ = — 2')s 
Yor Jia A(z Trthds 

where 2’, y', and z’ are the vertex of the conical surface. Here « 
and 6 vary with the position of the generatrix, Eliminate a, y and 
z between (484) and (485), and the result may be written 
(436) 8 = oa. 

Substitute into this for 6 and a, their values in (435), and we have 
for the equation of conical surfaces. 


(437) irr o(F=*). 


If ga be eliminated from (486) and (4385) by (429), we have, 
after clearing of fractions, 

(438) 3 —2' = p(x —z') + g(y —y’')- 

This is the relation among the partial differential coefficients of 
all conical surfaces. 

These two Propositions show the nature of the results obtained by 
eliminating an indeterminate function according to the process of 
Proposition LXXX. These results (in these two propositions) are 
deducible in a simple manner, independently of the Calculus. The 
first, (433qa), is the condition of parallelism of a line to a plane, as it 
is obtained in Analytical Geometry. The last, (438), is the equation 
of a tangent plane through the vertex, and designates that the tan- 
gent plane to a conical surface coincides with the surface along the 
generatrix, 

Many other Propositions are resolved by the process of elimina- 
tion indicated in Proposition LXXX ; but the foregoing will suffice. 


PROPOSITION LXXXII1. 
Determine the equation of the tangent line to a curve of 
double curvature. 
Let the equations of the projections of the curve of double cur- 
vature be, 
(439) sedi y= %, 
The equations of the tangent lines to these projections are as in 


Proposition III. | 
10° * 


166 DIFFERENTIAL CALCULUS. 


(440) ene ty ee 


These together represent the tangant line required, where 2’ yz’, 
are the variable co-ordinates. 

Ex,—Let the curve of double curvature be formed by the inter- 
section of the surfaces, 
(a) e+ P= y’, 
(5) xt+ y? + 2? = 2Rz, 
of which(a)is a cone whose vertical angle is 90°, and(b)is a sphere 
passing through the origin. The projections of their intersection are, 


(c) gt? = Ra, 
« circle, and 
(d) y? = Rz, 


a parabola. 
The tangents to (c) and (d) are the projections of the tangents to 
the line of intersection of the cone and sphere. 


PROPOSITION LXXXIV. 


Determine the surface formed by the tangents to a curve 
of double curvature. 


In Proposition LXVI, we observed that the surface formed by the 
characteristics of consecutive planes might be regarded as made up 
of the tangents to the Edge of Regression. The curve of double 
curvature may be regarded as the Edge of Regression of the sur- 
face formed by its tangents (440). Let the equations of the curve 
of double curvature be (439), and the equations of its tangent are, 


441 Pie Ori eee ae 
al Sear in tenn 


where o'z = —* =p, andyvz= “= 4. 
? a 1 vz rp Y 


The equations (441) contain the parameter z, which is the or- 
dinate of the point of tangency. Eliminate z between the equa- 
tions (441), and the result, which may be represented by 
(442) o(z', y', 7%) = 0, 
is a relation among the co-ordinates of any point on the tangent 


CURVES OF DOUBLE CURVATURE. 167 


line, irrespective of the point of tangency. Consequently, (442) is 
the locus of all the tangents to the curve of double curvature. 

Ex. 1.—Determine this locus when the curve is the intersection 
of the cone and sphere of last Proposition. 

Ex. 2.—Determine this locus when the curve of double curva- 


ture ls, 
(c) oh, set re y = bz? 
From (ec) we have, p = 2az, q = 2bz, and (441) becomes, 


(d) 2x’ —az? = 2ax(z'— 2), y' — be = 2bz (z' — 2). 
Multiply the first of (d) by 6 and the second by a and subtracting 

the results, we have, 

(e) mp b25 

which is the locus required. Equation (e) being the equation of a 

plane perpendicular to the plane XY shows that the curve (c) is a 

plane curve. And in general the result of the elimination of z be- 

tween equations (441) determines whether (439) is a plane curve or 

a curve of double curvature. 


PROPOSITION LXXXV. 

Determine the equation of the normal plane at a given 
point in a curve of double curvature. 

Let the curve of double curvature be represented by (489). If 
at any point in a curve of double curvature a normal line be drawn, 
this normal may take an indefinite number of positions, all how- 
ever, in the same plane. This plane is called the normal plane. 
[f 2’ y’ and z’, be any point on the normal plane, its equation, is 
of the form 
(443) gi — z= m (x'—x) +n (y' —Y)s 
where 2,y,z, are the normal point on the curve. 

Let (440) be the tangent line to (439). 

A plane perpendicular to a curve is perpendicular to the tangent 
to the curve at the normal point; and if a plane be perpendicular 
to a line, the traces of the plane are perpendicular to the projections 
of the line comparing the traces of (443) on ZX and ZY, with the 
projections of the tangent (440), we have, 


168 DIFFERENTIAL CALCULUS. 


ee een and ge oe ee end 
q dz Pp dz 
These values of m and n substituted into (448) give us for the 


equation of the normal plane, 
(444) fh mag tm a) ee (y ee 
dz dz 


By substituting into (444), the values of x and y, and their differ- 
entials from (439), the equation of the normal plane to curve (439) is, 
(445) a —2z + gz (2 — oz) + V2 (y' — 42) = 0. 

Ex.—Let the curve of double curvature be, 

(e) Gas, ye 02", 
find the normal plane. 


PROPOSITION LXXXVI. 


Determine the surface generated by the intersection of 
consecutive normal planes to a curve of double curvature. 

The normal plane (445), involves the parameter z, which is the 
ordinate of the normal point. Hence differentiate (445) for z vari- 
able. This differential, after dividing by dz, may be written, 


(446)  —1— (9'2)®—(a’z)*+ 9"2 (@' pz) 44's (y'—Ae) = 0. 
Eliminate the parameter z between (446) and (445), and we have, 
(447) 9(z',y',z') = 0, 


for the surface required. 
Equation (446) is the characteristic of plane (445). 


PROPOSITION LXXXVII. 


Determine the osculating plane at a given point in a curve 
of double curvature. 

Definition.—A curve of double curvature may be considered as 
a polygon of an indefinite number of sides. At a given point of this 
curve, two consecutive sides, and no more, lie in the same plane. 
The plane which contains any two consecutive sides is called, The 
Osculating Plane of the Curve of Double Curvature, 


CURVES OF DOUBLE CURVATURE. 169 


Regard, as in Proposition XXXVII., the middle points of each of 
two consecutive sides of the polygon as consecutive points. A tan- 
gent line to the curve may be regarded as the prolongation of one 
of its indefinitely small sides. 

Let 2',y',z’ be the given point on the curve of double curvature. 

Let the equation of the curve be represented by 


(448) ee Os, y= Ae. 
The equation of any plane through 2’,y',z’ is of the form, 
(449) z—s3 = m(r—-x)+n(y—y’'), 


where z, y, and z are any points on the plane. 
The equation of the tangent line to (448) through 2',y’,2’, is 
Ae ¥ a he Sy 
(450) Rene Fi wha #) and y irae ake z'). 
In order that the line (450) may coincide with plane (449), we 
have, from the principle of coincidence of a line and plane, [Analy- 
tical Geometry ], 
(451) m ie +n oe 
dz' dz' 
For the coincidence of plane (449) with the tangent line to (448), 
through any other point 2'’,y'’,z’’, we have in like manner, 
& dx'' dy" 
(452) m zr +n PD 
When the points 2',y’,z', and «'’,y'’,z’’ are consecutive, we have, 
Ce ae as Ye = 9 dy. and. oa 42 eae. 
trom which, regarding z' as the independent variable, we have, by 
differentiation, 
(453) da" = dz'4+ d@2', dy" = dy'+ d’y’, and dz"= dz’. 
Substitute (453) into (452), and subtracting (451) from the result, 
we have, 


—l1-—o, 


—l=+o. 


Ox! d’y' 
454 mm —— a Ce 
( ) dz" ae 2 dz" 

The two relations (454) and (451) suffice to determine m and n, 
and their values put into (449) make known the osculating plane. 
If we put 

“Se elimi ie da’ dvy’ _ 
(455) ee P; germ q; rai r, and ogi t, 


. 
170 DIFFERENTIAL CALCULUS, 


and solve (454) and (451) for m and n, we have, 
(456) Th os and Myf soe pt Mahala 
pt — qr pt — qr 
and hence (449) becomes, 
(457) (pt — qr) (z—2') — t(a@—a2') + r(iy—y)=o0, 
which is the osculating plane required. 
Ex.—Determine the osculating plane to the curve, 


ae, y = (R® — 2’)'. 


PROPOSITION LXXXVIII. 


Determine the radius of curvature at a given point on a 
curve of double curvature. 


To a curve, considered in space, different osculating circles 
may be applied at the same point. The plane of the osculating 
circle, whose radius of curvature we here require, coincides with the 
osculating plane (457), determined in last Proposition. The radius 
of curvature must also lie in the normal plane to the curve of 
double curvature. Hence, it coincides with the intersection of the 
normal and osculating planes, and as the osculating circle must co- 
incide with the curve at two consecutive points, (Prop. XXXVIII.), 
the centre of curvature must be on the characteristic or line of in- 
tersection of the normal plane with its consecutive plane. 

Adopting the notation (455), the normal plane (444) may be writ- 
ten (2z',y',z', being the normal point), 

(458) G—2 + p(x—2')+qiy—y')= a, 
and its differential corresponding to (446), is 
(459) —(l+p4+q@)+r(¢—z') + tyy¥—y)=o0. 

If x, y, and z, be the centre of the osculating circle, the radius 
of curvature calling it R, is 
(460) R= (@— a + (y—y' 4 ( — ap, 

At the centre of curvature x, y, andz,arecommon to the osculat- 
ing plane (457), to the normal plane (458,) to the characteristic 
(459), and to the radius of curvature (460). Hence, eliminate 2, y, 


CURVES OF DOUBLE CURVATURE. 171 


and z,-from (460) by means of (457), (458) and (459), and we have 
the radius of curvature required. The result of this elimination is, 


2\3 
(461) etc la Se ne Pa 
e+ r+ (pt — qzy 
Ex.—Determine the radius of curvature of the curve, 
x= az’, y = bz + 2. 


PROPOSITION LXXXIX. 


Determine the centre and radius of curvature of an oscu- 
lating sphere. 

Definition. 

An osculating sphere to a curve of double curvature, is one whose 
surface contains three consecutive points of the curve. 

Let the curve of double curvature be, 

(462) y = @2', and 2z'= yz. 

Let P,P’, P’’ be three consecutive points on (462), and let (a,8,c) 
be the centre of the sphere required. The equation of the sphere 
is of the form, 

(463) R? = (a— 2)? 4- (8 — yf + (C—7/. 

Suppose the centre (a,3,c) be on the normal plane to (462). 
The equation of the normal plane (458) is, for the centre of the 
sphere, 

(464) e—2z +p(a—2’')+q(B— UO 

The characteristic (459) of this normal plane, is 
(465) —(l+p+/)+r(a—r)+t(@—y) =o 

Differentiate (465), and we have for the intersection of (469), 
with its consecutive characteristic, 

(466) (a—a')r' + (8 —y') t —3 (pr + gt) = 9, 
where eye e and *es Gi 

Now, if the sphere whose centre is at the intersection of (466) of 
two consecutive characteristics, passes through one point P on (462), 
it also passes through the three consecutive points P, P’, P’ on 
(462): for the two consecutive characteristics are on three consecu- 


—— a 


172 DIFFERENTIAL CALCULUS. 


tive normal planes, which pass through P, P’, P’. The sphere 
whose centre is at the point (466), is therefore, the sphere required. 
The four equations (463), (464), (465) and (466), make known the 
values of a,6,c, and R, which determine the centre, and radius of 
the osculating sphere. 

Cor. The locus of the centre of the osculating sphere is the Edge 
of Regression of the envelope of the normal plane. For the equa- 
tions (464), (465), (466), are those employed in determining the en- 
velope and Edge of Regression. [See Propositions LXXXVI, &c.] 


PROPOSITION XC. 


Determine the evolute of a curve of double curvature. 

If a cord be wrapped round the evolute of a plane curve, and 
continued to a point in the involute, the extremity of this cord will, in 
unwinding, describe the involute, [Proposition XL.] 

The involute of a plane curve may likewise be described by the 
extremity of a cord unwound from a right cylinder, whose directrix 
is the evolute. 


For let BM be the involute 
whose evolute in the same plane 
is CD. Through the consecutive 
points E,F,G, &c. erect lines EL, 
FK, GH, &c. perpendicular to 
the plane CD. These perpendi- 
culars will form the envelope, or 
cylindrical surface that would be 
produced by the characteristics 
of normal planes to the curve 
BM. Let HGM be one of these 
normal planes, HG its charac- M 
teristic, and GM its trace on the plane of CD and BM. 

Suppose from any point H on GH, a line HM be drawn in the 
normal plane HGM to the involute at M. If we now suppose the 
normal plane HGM to be wrapped round the envelope HGEL, the 


FIG. 71. 


CURVES OF DOUBLE CURVATURE. 173 


line GM will be wrapped round the evolute CD, and its extremity M 
will describe the involute MB; the line HM will be wrapped round 
the envelope HGEL, and its extremity M will likewise describe the 
involute MB. The line HM in this process will form on the enve- 
lope a curve HKL, which may also be regarded as the evolute of 
the curve MB. Hence any involute MB has an indefinite number 
of evolutes HKL, all traced on the envelope HGEL. 

[t is obvious that these evolutes will become straight lines when 
the envelope HGEL is developed or rolled out on.a plane. Any line 
MH is tangent to the evolute HKL, for MH is the prolongation of 
Aik, one of the indefinitely small sides of the polygon that com- 
poses the evolute. 

If the curve BM be a curve of double curvature, and HGEL be 
the envelope of the normal plane to BM, take H any point on this 
envelope, draw HM in the normal plane to M, and suppose the nor- 
mal plane HGM to be wrapped round the envelope, the point M will, 
as in the plane curve, describe the curve of double curvature MB. 
The curve HKL, traced on the envelope by the wrapping of the 
cord HM, will be the evolute of the curve MB. The evolute HKL 
will, as in the plane curve, have HM for its tangent, and become a 
straight line when the envelope HGEL is developed. There may 
obviously be an indefinite number of evolutes to BM, the curve of 
double curvature, all traced on the envelope HGEL. 

To find the equation of any one of these evolutes, let x',y’,z’ be 
any point M on the given curve of double curvature, and L,Y,% any 
point Hon the evolute required. Let the curve MB be repre- 
sented by 


(467) y = 92’, x! = Ve", 
The equation of the normal plane HGM is (458). 
(468) i—2 + p(@—2')+q (y—y’). 


Since MH is tangent to the evolute HL, the equation of MH is 
for the point M, 
(469) z2—r= a (z' — 2), 


xy ‘ d 
(470) y— y= = (2 — 2). 


174 DIFFERENTIAL CALCULUS. 


Let the equation of the envelope HGEL, as found in Proposition 
LXXXVI. be, 
(471) $(2,y,%) = 0 

For the point M, the co-ordinates x', y', and 2’ are common to 
(467), (468), and (469) ; and for the point H we have 2, y, and z 
common to (468), (469), and (471). Hence eliminate 2’,y’,z' 
from (467), (468), (469). The result may be represented by 


(472) nf Z,Y,%. e) =O. 


If y be eliminated between (472) and (471), the result, which 
may be written, 


dx 
473 Ceo ae EN, \ pe — SO 
(473) +f zy 0, 


will be the differential equation of the projection of the evolute on 
the plane of XZ. . 

By the aid of the Integral Calculus, we obtain from (473), the 
projection of the evolute on XZ, which with surface (471), deter- 
mines the evolute in space. 

If G be a point where the osculating plane (457) to the curve at 
M, cuts the envelope HGEL, then is MG shorter than any other 
radius of curvature HM to the curve at M, i. e. MG is perpendicular 
to GH. This shortest radius is known by the name of The Abso- 
lute Radius of Curvature, and is the one whose length is deter- 
mined at (461). 

One property of these absolute radii of curvature is, that no two 
of them are in the same plane. Therefore, they do not intersect, 
consequently, the locus of the centres of the absolute radii of cur- 
vature is not an evolute of the curve. 


PROPOSITION XCI. 


A tangent is drawn to a curve of double curvature; at 
the point of tangency a line is drawn, making a given angle 
with the tangent, find the locus of the intersection of this 
line with its consecutive line. 


CURVES OF DOUBLE CURVATURE. ¥75 


FIG. 72. 


Let the curve of double curvature 
be represented by 
(474) y= 9%, © 2, 

Let its projection on zx be PD, 

Let PR be the projection of the 
tangent line. 

Let PB be the projection of the 
line. ya 

The equation of the tangent line is, 
475) Y—y= 2 (Z—», X-2= 2 (7 —», 
where x, y, and z are the co-ordinates of the point of tangency and 
the equation of the line making the given angle with the tangent, is 
(putting x’,y’, z’ for its co-ordinates), 
(476) y —y=d(2z' —2), x! —— 2% = a (z',— 2) 

Put C for the cosine of the angle whose projection is BPR and 
[by Analytical Geometry,] we have for the angle included between 
the lines (475) and (476). 


lx dy 
| iad a Gee 

yee 5 dz vi dz’ 

(477) OE meee | 1 pes 5 re 
Lote at Be GL te ete he 
yak ) y dz? dz? 

dx dy 


By means of (474) the differentials, ae in and x and y may 


2° dz 

be eliminated from (477), and the result solved for z may be repre- 
sented by 
(478) z= 9(a, b), 
which, of course, includes C and other constants. 

Differentiate (476) for «',y,'z' constant, and a and b variable, and 
eliminating x’ — z between the two differentials, we have, 
(479) dx: Zio add Aix: ad 21bdan 

dz dz 

which is the relation among the differentials when the line (476) in- 
tersects its consecutive line 


176 DIFFERENTIAL CALCULUS. 


By means of (474) we can eliminate, 
dx dy 
dz dz 
from (479), and then by means of (478) we can eliminate x from 
the result, and (479) may then be written, 
(480) db 9'(a,b) — da y'(a,b) = adb — bda. 

Here again we must have recourse to the Integral Calculus, in 
order to obtain the equation whose differential is (480). 

This original equation can contain no other variables than a 
and 6. If we suppose this original equation to be obtained, and to 
be represented by 
(481) big. fi, 
we may solve (478) and (481) for a and 6. 

The result may be represented by 
(482) b = Fz, and! ai=*fz: 

By means of (482) and (474), the equation (476) of the line 
becomes 
(488) y'’ — oz = F2a(z'— 2), and 2 — w= fz (z' —2). 

Differentiate either of the equations (483) for z variable, and 
eliminating z between this differential and each of the equa- 
tions (483), we will have the projections of the curve required, 
which may be represented by 
(484) a a and # = fine 

Cor. If z be eliminated between the two equations (488), the re- 
sult which may be represented by 
(485) F(z’, y'. 2) = 0, 
is the developable surface which is the locus of the line (483), and 
of which (484) is the Edge of Regression. 


PROPOSITION XCI. 


A tangent line is drawn to acurve of double curvature, a 
right cone of given vertical angle has its vertex at the point 
of tangency, and the tangent line for its axis, determine the 
locus of the intersection of this cone with its consecutive 
cone. 


TWISTED SURFACES, 177 


Let the line (476) be the generatrix of the cone, and C being the 
cosine of half the vertical angle, we have as before, (477). If we 
substitute into (477) for a and 8, their values from (476), the result 
is the equation of the cone, containing the three co-ordinates of the 
vertex. By means of (474), two of these co-ordinates, as a and Y; 
may be eliminated, and we have the equation of the cone, which 
may be represented by 
(486) o(2',Y',2', w)= 0. 

Eliminate the parameter z between (486) and its differential for 
z, and we have the surface required, which may be represented by 
(487) FP’ (x’,y',z') = 0. 

Cor, 1.—Surface (487) contains all the lines (484). 

Cor. 2,—If half the vertical angle is a right angle, then C = 0, 
and (477) becomes, 


. la di 
488 1 A Bae 
( ) it Ye * dz ‘ 


and the cone becomes the normal plane, and the Proposition becomes 
Proposition LXXXVI., which is a particular case of the present. 


PROPOSITION XCIII. 


Determine the equation of a twisted surface. 

A twisted surface is one generated by a straight line moving in a 
given manner, and continually changing the plane of its motion. 

Let the equation of the generating line be, 

(489) : =az + Jb, 
y= cz+e, 

The four parameters a,b,c,e, which enter into (489), evidently 
change when the generatrix changes its position. If three of these 
parameters, as b,c,e, can be expressed in terms of the fourth, a, the 
parameter @ can be eliminated between the two equations (489), and 
the result will be the surface required. 

To express b, c, and e as functions of a, requires that the line 
(489) be subjected to three conditions. 

The three conditions to which the generatrix (489) may be sub- 


jected, are either that it shall touch three given lines, or that it shall 
17 * 


178 DIFFERENTIAL CALCULUS. 


touch two given lines, and be parallel to a given plane, This divides 
the Proposition into two cases. 


Case 1st. 
Let the generatrix (489) touch three given lines, Let the three 
lines be denoted by 


(490) piety AF x = Fz, 
(491) y =f 2, x= F's, 
(492) Yosef 25 tad ame 


Where the line (489) touches (490), the x, y, and z are common 
to the four equations (489) and (490). Eliminate the three co-ordi- 
nates x, y, and z between these four equations. We may represent 
the result by 
(493) 9(a,b,c,e) = 0. 

In like manner, and for a like reason, eliminate the three co-ordi- 
nates between the four equations (489) and (491), and we may 
represent the result by 
(494) o'(a,b,c,e) = 0. 

Again, eliminate the three co-ordinates between (489) and (492), 
and represent the result by 
(495) p''(4,b,c,e) = 0. 

Solve (493), (494), and (495), for b, c, and e, and represent the 
results by 
(496) P= pds Coin, e= na, 

Substitute the values of 6, c, and e, from (496) into the genera- 
trix (489), and we have, 

(497) i = az + oa, 
y = jaz + xa. 

Eliminate a@ between the two equations (497), and we have, 
(498) o(X,y,2) = 0, 
the surface required. 

The lines (490), (491), (492), may be either straight lines, or 
curved lines, or some of them straight and others curved. 

Ex.—Let the three lines on which the generatrix moves be the 
three straight lines, 


TWISTED SURFACES. 179 


t= 0 y = 0, 
x= M2, y=nz + 4q, 
x= mz + Pp, y = nz 
Determine the generatrix (497). 
Case 2d. 


Let the generatrix (489) touch two given lines, and be parallel to 
a given plane. 

Let (490) and (491) be the two given lines, and let the given 
plane be 
(499) z= me + ny + B. 

Since the generatrix (489) is to be parallel to plane (499), we 
have, by the conditions of this parallelism, [Analytical Geometry], 
(500) ma+ne—1l=o0. 

Derive (493) and (494) as before. Solve the three equations 
(493), (494), and (500), for b,c,e. Represent the results by (496), 
and we have as before, (497) for the directrix involving one para- 
meter, and (498) for the surface, 

Ex.—Let the two lines on which the generatrix moves be one the 
axis of y, and the other, 

(A) sms Nn, and Tne 

and let the plane to which the generatrix is parallel, be the plane of 
XZ. The equation of the generatrix, on the axis of Y, and parallel 
to XZ, is 


(B) y =a az, Ys ¢, 
Eliminating 2,y,z between (A) and (B), we have, 
(C) ae = me + pn. 


Eliminate a and e between (B) and (C), and we have for the sur- 
face required, 
yx = may + Nps. 
Ex. 2.—Find the twisted surface generated by a line moving par- 
allel to ZX, one end of it on the axis of Y, and the other on the curve, 
y” = Amz, ag ( £ 


By the procedure of Proposition LXXX., one or all of the func- 
tions of a may beeliminated from (497). 


APPENDIX 


DIFFERENTIAL CALCULUS. 


PROPOSITION A. 
Determine the circle which touches three given curves. 
Let the equation of the circle be 


(1) (x — a)? + (y— a” = BY 
and let the three curves be represented by 
(2) y = 9x, 
(3) y" a o" Cae 
(4) fe, he: a 
Where (1) touches (2), (3), (4), respectively, it becomes 
(5) (c’' —a)? + (y' —fsy = B, 
(6) (2 — a) + (y” — 8) = RY, 
(7) (zt ee a)? fe (yf ais BY == R?2, 


Let p be the differential coefficient of the circle (5), (6), or (7), 
and p’, p”, p’’, the differential coefficients of (2), (3), (4), re- 
spectively. Then, since the circle touches each of the curves (2), 
(3), (4), we have, 


(8) P=P 
(9) p=P'; 
(10) P= Lae 


Eliminate 2’, y’, from (2), (5), and (8), and we may represent 


the result by 


(11) F (a, 8, R) = 0. 

Eliminate x" and y'' from (3), (6), and (9), and we have, 
(12) F’ (a, 6, R) = 0. 

Eliminate x’, y'” from (4), (7), and (10), and we have, 
(13) F” (a, 8, R) = 0. 


(181) 


182 DIFFERENTIAL CALCULUS. 


Solve the three equations (11), (12), (13), for a, 6, and R, and 
these values determine the circle in position and magnitude. 

This Proposition shows how the Calculus may be applied to 
Geometrical propositions in which the curve (in this case the circle), 
is limited to one curve of the species. 

Cor. If the circle pass through one point (m,n,) and touch two 
curves, viz. (2) and (3), equation (7) would become 
(14) (m — a) + (n— py = R’, 
which takes the place of one of the relations, (11), (12), (13). In 
a similar manner, if the circle passes through two points, and 
touches one curve, the condition of passing through two points, 
furnishes two of the equations (11), (12), (18). The condition of 
touching a given curve (2) furnishes the other, and the circle is de- 
termined. 

We will generalise this process in the next Proposition. 


PROPOSITION B. 


A curve containing a given number of parameters touches 
as many curves as it has parameters, determine the curve. 


It has already been observed that a curve or surface may be 
subjected to as many conditions as it contains parameters. Hence, 
a curve containing n parameters, is fixed by subjecting it to n condi- 
tions—such as touching n curves—or passing through n points—or 
passing through m points, and touching ~— m lines or touching 
n—m curves, and having m parameters of a given magnitude, &c. 
Hence, if the required curve has % parameters and touches n curves, 
each of the differential coefficients of the n curves is equal to the 
differential coefficient of the required curve. This furnishes x 
equations. ‘The m curves furnish m more equations, and the re- 
quired curve taken 7 times, i.e. once for each point of tangency 
furnishes 2 more equations. Hence, we have 8 equations, from 
which if we eliminate the 2n co-ordinates, we have n equations 
among the n parameters, from which they may be determined, and 
this fixes the proposed curve. 


APPENDIX. 1838 


Ex.—Let the curve be the ellipse. 

The equation of the ellipse in its general form is 
(15) a’? (y—n)y + B(x —m)* = a’ B’. 

This contains four parameters, a, b, m and n. Hence, it may be 
subjected to touch four given curves. Let 
(16) y = fe, y =f 2, y = fix, y=foo, 
be the four curves to which (15) is tangent, 

Then, if p be the differential coefficient of (15), and p',p",p'""»,p'""”’, 
be the differential coefficients of (16) respectively, we have, 

(17) p=p,* p=p"', p=p", p=p'- 

Since (15) touches each of the curves (16), we may take (15) as 
four different equations, according as it touches the first, second, 
third, or fourth of (16). Hence (15), (16), (17) are twelve equa- 
tions, from which we may eliminate the eight co-ordinates of the 
points of tangency, and we will have as the result, four equations, 
among the parameters a,b,m,n. By solving these four equations, 
these parameters may be determined, and the ellipse (15) described, 


so as to touch the four curves (16). 


PROPOSITION C. 


Determine the equation of the cycloid. 
Definition.—If a circle roll along a given straight line AB, the 


curve described by a point C on the circle, is called the cycloid. 
FIG, 73. 


Let the point C 
coincide with the 
given line AB at A. 

Take A as the 
origin, and the giv- 
en line AB as the 
axis of X. Let AR x B 
x = OD, the radius of the generating circle. Then it is obvious 
that when the circle rolls along AB, every point in the circumfer- 
ference passes in contact with AB. Consequently AD = CD. 

Let (z,y) be the point C, then 


184 DIFFERENTIAL CALCULUS. 


(18) AF = AD — FD, 

and we obviously have, 

(19) AD = CD = versin.~'y, and 
(20) FD = CE = (2ry —y)?. 


These values put into (18), we have, 


(21) © = versin.~'y — (2ry — y?)?. 

This is the equation of the cycloid. 

If we differentiate (21), we have, after reducing the second side 
to a common denominator, 
(22) dx =e yy < 
(2ry — y’)" 

By means of (21), or (22), the other properties of the curve can 
be determined. 


PROPOSITION D. 

Determine the angle which the radius vector makes with 
a given curve at a given point. 

FIG, 74, 

The angle which a line makes A 
with a curve at any point is the 
same as the angle which the line 
makes with the tangent to the 
curve at that point. i A D 

Let PR be the given curve, and P the given point on it. 

Let the origin A of rectangular axes be the pole, and the axis of 
X the angular axis. Let 

fo edits Ps PAX = Q, AD \=4 Lr, PD = y. 

Draw PL tangent to the curve at P, then LPA is the angle re- 

quired. By (57), Differential Calculus, we have, 


dy 
a — = tan.L = p, 
(2) FP P; 
and by the figure we have, 
() tano = 2. 


APPENDIX. 184 


Also by the figure we have, 
(c) tan.LPA = tan.(PAX — PLX), 

Expand (c), and substitute into the expansion the values of the 
angles in (a) and (b) and we have, 


(d) tan, LPA) = 94% 7 dy 
xdx + ydy 
But in the figure we have, 
(e) L = p COS.a, yy = p SiN. 
Differentiate (e), and eliminate x,y, dx, and dy, from (d), by 


means of (e) and the differentials of (e), and we have, 


(23) tan.LPA — f4e 
dp 
This is the value of the tangent of the angle required. 
If the equation of PR be 
(f) p = Po. 
we put the value of do ~ dp, deduced from (f ), into (23) and we 
have the tangent of the angle in terms of the polar co-ordinates. 
By means of the trigonometrical relation, 
(24) tan, =) SA 
COs. 
we can, by using the values of the tangent of the angle in (23), 


determine the values of the sine, cosine, secant, &c., of the angle 
LPA? ViZ. 


> 


(25 sin. LRA, eet Oy 
(dp? + pdw*)? 
(26) cos.LPA = dp 


= % 
(dp? + p dw*)* 
Ex.—Let the curve PR be p = aw”, find the sine, cosine, and tan- 
gent of the angle LPA. 


PROPOSITION E. 
SUPPLEMENTARY TO PROPOSITION XXXVII. 
Determine the locus of the intersection of two consecutive 


lines. 
18 


186 DIFFERENTIAL CALCULUS. 
FIG. 75 


Let the equation of MOG be 
(27) (2,y,8) = 0 = u, 

This line, [whether straight 
or curved], evidently depends 
for its position on the parameter 
8, which enters its equation 
(27). If we differentiate (27) 
for x and y variable, and 6 con- 
stant, we have, 


(28) du dx 
dx 
If we differentiate it for y and @ variable, and a constant, we have, 
é du dy du dg 
29 Es rls. i eee 
(29) ay tae P 0 


The differential (28) may be regarded ag the passing from a point 
C to the consecutive point R on the same curve, and (29) as the 
passing on the ordinate EB from a point C to a point B on the con- 
secutive curve. It is obvious at the point O, where the consecutive 
curves intersect that both x and y remain constant in (27), while 
g varies. Hence, if y be constant in (29), dy is zero and that equa- 
tion becomes for the point O, 

du ds 


(30) ee en Oy 
dp 
and we have (27) and (30), to eliminate 8, the result will be of 
the form, 
(31) 5 red Be 


and denote the locus of the point of intersection. 

Cor. The curve (27) is tangent to (31). 

For the x and y of (27) are the x and y of (31) at the point of 
intersection O, and if (27) be differentiated for a, y, and 6 variable, 
we have, 
(32) G0 eee Ce ae 

dx dy dg 
which, when £ is determined by (30), is the differential equation of 


> 


APPENDIX. 187 


(31). But the condition (30) reduces (382) to (28). Hence, the 
differential coefficients of (27) and (31) being the same, the curves 
are tangent to each other. 


PROPOSITION F. 


Determine the length of the polar subtangent of a given 


curve. 
FIG. 76. 
Draw PH tangent to the curve P 


at P. Let A be the pole. 


Draw AH perpendicular to 
: H 
the radius vector AP. R 


In the right angled triangle 
PAH, we have, by Trigonometry, L A D 
(33) AH = AP. tan. HPA. 

Put p for the radius vector AP, and putting into (33) for tan. 
HPA, its value at (23), Proposition D we have, 
(34) INE a ay 
dp 
which agrees with the result in Proposition X XXIII. 

3y means of Propositions D and F, we deduce the values of the 
polar subtangent, subnormal, &c., without having recourse to the con- 
sideration of indefinitely small quantities. 


THE 


INTEGRAL CALCULUS. 


CHAPTER I. 


PRINCIPLES OF INTEGRATION. 


Tse Integral Calculus is the inverse of the Differential, and 
teaches the method of returning from a given differential to the pri- 
mitive function from which it is derived. This process is termed 
Integration. ‘The Rules of procedure in this Calculus are therefore 
to be obtained by inverting those of the Differential Calculus. 

The first rule of the Differential Calculus teaches the method of 
differentiating a monomial raised toa power. The inverse of this 
is the first rule of the Integral Calculus. 


RULE I. 
To integrate a monomial, increase the exponent by unit , divide 
Ee P y Af 
by this increased exponent, and by the differential of the variable. 


m 4-1 


» which is made 


The integral of xdz is, by this rule, 
m+ i 


by increasing the exponent by one, and dividing by m +. 1, the in- 
creased exponent, and by dz. The differentiation of this last expres- 
sion produces the first. Hence we may say, generally, that the 


object of integration is to obtain an expression whose differential is 
the differential proposed. 


The symbol f is employed as the reverse of the character 


d. It is called the sign of integration, and denotes that the quan- 


tity before which it is placed is to be integrated, Thus / 2"dx 
5 18 * (189) 


196 INTEGRAL CALCULUS. 


denotes that x"dz is to be integrated. This symbol is the first letter 
of the word sum, and was used by the first writers on the Calculus, 
to express that integration was the summing, or adding together of 
the indefinitely small quantities of which the integrated expression 
is composed. Thus if y represent a straight line, dy is an indefi- 
nitely small portion of that line, and to integrate dy is to sum or add 
together these small portions. Hence the integral of dy is found by 
simply removing the characteristic d from dy. 

According to Rule II. of the Differential Calculus, a constant 
quantity added to a given function disappears in differentiation. 

Inversely we have, 


RULE IL. 
In integrating, a constant quantity must be added to the integral. 
Thus the complete integral of 2"dz is, 
co 


where C designates the constant that may have disappeared in 
differentiation. ‘The method of determining this constant will be 
given hereafter. 

By Rule III. of the Differential Calculus, in differentiating the 
product of a constant factor and a variable, the constant factor 
remains. Inversely we have, 


+ C, 


RULE Ill. 

In integrating, if there be a constant factor to the differential, it 
remains. 

Consequently, in such an expression as aa"dx, we may put the 
factor a before the sign of integration. Thus, 
res foie) 
n+l 

By Rule IV. of the Differential Calculus, the differential of the 
sum of any number of variables is the sum of their differentials, 
Inversely we haye, 


(2) fwd are fuvde oe 


RULE IV. 


The integral of the sum of any number of differentials is found 


PRINCIPLES OF INTEGRATION, 19] 


by integrating each term separately, and adding the terms to- 


gether, 
Thus if we have, 
(3) dy = ax*dx + bx dx — r'dz, 
the integral found by integrating each term is, 
ax® bx? a 
4 a = — yl IP pegs 
) / 3 F 2 5 


These Rules apply to the cases where each term to be integrated 
contains one variable, and no more. For sake of convenience, we 
will sometimes put our examples in the form of equations, taking 
only that form of equation in which the first side is a simple 


differential. 
As examples under the foregoing Rules, take the following. 


1 pba s 4 
dy =adx + az7dr_ .-. = ar + ri 2? + C. 
dz a 
dy © = ax de>. et (se MER 
z + s 32° 
tan 
ade = dex 7 a ee th, noeat C 
Be Siyiace wie ti oA Sh de eae Y 


dy = ax™ dx + aida + dz. 


In these examples, we observe that if the variable in the monomial 
be in the denominator, it must, (with an exception to be presently 
noticed), be brought to the numerator before integrating. 

In the Differential Calculus, (19), we observed, that a polyno- 
mial, as 
(5) = (a mia bx*)", 
could be differentiated as a monomial by regarding the part within 
the vinculum as a single quantity raised to the power n. The dif- 
ferential of (5) is 
(6) dy = n(a + ba*)"' 4 bz? dz. 

To produce (5) from (6), we must obviously increase the ex- 
ponent of the binomial by unity, divide by this increased exponent 
n, and by 4bz* dx, which is the differential of the part within the 


vinculum. Hence, 


INTEGRAL CALCULUS. 


RULE V. 


A polynomial may be integrated as a monomial if the part with- 
out the vinculum be the differential of the part within. 

In the two following examples the part without the vinculum is 
the complete differential of the part within. 


Ex... dy = (a + ba)" boa. Mie tc (Cee os 
n+l 


to|— 


Z 
Ex.2. dy=(a+b2*) 2badx -.y=2(a + ba*) +C. 
If we had such an expression to integrate as, 

(7) dy = (a + br)” dz, 

where the part without the vinculum is the differential of the part 

within, except by a constant factor b, we may multiply by 6 the 

factor required, then integrate, and divide the integral by the factor. 

The integral of (7) is, by this process, 


bx)” +1 

(8) d b(n + 1) a 
It is usual to- denote the integral of (7), thus 
(9) — f (a + bx) bde, 


where, by multiplying the part after the sign of integration by 6, 
and putting the reciprocal of 6 before the sign of integration, the 
expression remains unchanged in value, and the part outside the 
vinculum is the complete differential of the part within. If the 
part outside the vinculum requires a variable factor to make it the 
differential of the part within, the binomial cannot be integrated by 
the foregoing process. Thus, if we have, 
(10) dy = (a + bzx*)” zdz, 
a variable factor is required to make the part without the vincu- 
lum, the differential of the part within, and (10) is not immediately 
integrable by Rule V. 

If we have such an expression as, 


(11) ay = coi dy = xdz. 
es 


PRINCIPLES OF INTEGRATION. 193 


If this be integrated by Rule I, we have, -. y = infinity, and 
the integration fails. 

But the numerator of (11) is the differential of the denominator, 
and by Rule V of the Differential Calculus, we know that this came 
from differentiating the logarithm of the denominator. The integral 
of (11) is, therefore, 

(12) ee log.a + C. 

Hence, we have, 


RULE VI. 


If the numerator be the differential of the denominator, the in- 
tegral is the logarithm of the denominator. 


dz , 
Ex. lle em  y = log. (a + zt) + C. 
Ex, 2. dy = tks. Sass “- y = log. (a + bz’) + C, 
a+ b2? 


If the numerator fail by a constant factor of being the differential 
of the denominator, multiply by this factor, then integrate, and 
divide the integral by the factor. If, for example, we have, 


(13) dy.— 


a+ bx 

here the numerator, is the differential of the denominator, except 
by the constant factor 6. ‘Then, multiplying by this factor 8, in- 
tegrating and dividing by 8, we have, 


(14) *) i= = log. (a + bx) + C, 
As another example, take 
dy — _%4 
Ar a— x’ 


here the numerator wants the factor—2 to be the differential of the 
denominator. Hence, the integral is 


y = — + log. (a—2*) + ©. 


If we put C = log. c, (12) becomes 
(15) y = log.x + log.c = log.cz. 


194 INTEGRAL CALCULUS. 


If e be the number whose log. is unity, (15) may be written 
(16) pores OF, 
for the log. of (16) is (15). 

These essential principles will be employed hereafter. 

As an illustration, take the following examples. 


x'dx 1 
oo o*. = —_— — ] 3 —b ¥ bs 
ay a—bx* 4 cy Me gia he 
rdx 1 
dy = ooraias y= aT? log.(a + ba®) + C, or 
= = log.c (a + bz’), or, 
ev = (ca +- cba!) =. 
n—1 ; 
Pall Gps eal lb td aed 
tne ax + x 


INTEGRATION BY CIRCULAR ARCS. 


We have seen in Differential Calculus, Proposition XXXII., that 


if we have, 
(17) y- = sin-'y, 
then by differentiating we have, 
(18) ahi 
persia 
Consequently (17) is the integral of (18), i. e., 
(19) AP es = sine. 
(1-2?)! 
If we had the form, 
adx 


(B aia cx)? : 
then putting c = bm, it becomes, 
a mdx 


bm (l— mx)" 


PRINCIPLES OF INTEGRATION. 195 


and the integral is, 


(20) = f oe = — sin—mz + C, 
bm (1 a m*x”)* bm 
or restoring the value of m, 
(21) af ssa = sing! — 4.0, 
(b? a cx*)t c b 


This will serve as a form for integrating all similar expressions. 
We observe in (20) that the sine line is ma, and the radius unity. 
Again, by Differential Calculus, Proposition XXXII., 
— dx 


(22) d.cos.—! 6 pee Pe FER al 
(1 — 2’) 


Consequently, 


(23) {ee == (0s. - me 
(1 — 2’): 


If we had the form, 
— adx 
(b eer c?x?)8 . 
putting c = bm, then integrating and restoring the value of m, 
we have, 


(24) cf Poi safe aCOstt te, C. 
: (0 Ait cx?)* c b 
By the same Proposition of the Differential Calculus, we have, 
(25) d.tan.—"u = te 
1+w 
Integrating, we have, 
(26) id HE tan.—'w, 
1 + uv 
If we had the form, 
a du 
B+ cu? 


then putting c = mb, this reduces to 
a m du 
bm dL + om? 


i — ee = 
a area. ae -_— 
<a ae 3 Ge 


+ eee 


t 


196 INTEGRAL CALCULUS, 


Since mu may be regarded as the tangent line, this may be inte- 
grated by comparison to (26), and we have, 


(27) ve f peste Bala ONT APs tan.—’mu. 
b?m 1 + mu b’m 
Restoring the value of m, (27) becomes, 
du a cu 
28 aff BOF Soc. 2 sane SY eee 
) Pte bb 
which will serve as a form for integrating all similar expressions. 
As an example, take the equation. 
lx 
d ij CE 
(d) = pS 
Comparing this with (28), we have b —2, c=1, and (d) 
becomes, 
1 x 
€ °*. = — tan’ — + C. 
(€) y= ane 
Again, by Proposition XXXII., Differential Calculus, we have 
(29) d.versin.—y — OB. 
(20 — v®)# 
Consequently, 
(30) f caw, , = versin,—'v. 
(20 — v?)? 
If we had the form, 
(31) dy = sir ‘ 
(bx Rana. cx") 
rT aa o> Pp 
assume “pO may ami 
Put this value of x into (31), and it becomes, 


ce (20—-) 
and integrating by (30), we have, 
(33) a versin.—'» +. C. 
c 
Restoring in this the value of v, we have, 
2 
(34) f 1 Of a eevee. 70% Zeer 


(bx — c*x?)* Cc 6° 


RATIONAL FRACTIONS, 197 


We can integrate many expressions by the form (84). 

As an example, take the following : 

: adx 
Sf) te 
(oo msx*)* 

Comparing this with (34), we have, b®? = 3, c? = m, and (f) 
becomes, 


. Q2mx Y 
y= © yersin. ea oe 
vm = 
Integrate the expressions 
dy — oe dy = Me ale dy et aj atte. 
(3—z’)" 2 +. 52" (72 — 5x”): 


CHAPTER II. 
INTEGRATION OF RATIONAL FRACTIONS. 


A rational fraction is one that contains no surd quantity. 


Lemma. 

Divide a rational fraction into several partial fractions. 

This is a procedure of Algebra; but as it is not much treated of 
in books on that subject, it may be well to explain it here. Let the 
fraction be 

; ax 
(39) ghd Eo BS i 

Any such fraction can be divided into as many partial fractions 
as there are factors in its denominator. To find these factors, put 
the denominator equal to zero, that is, assume the equation 
(36) x?— 22 —3= 0, 

Find the roots of this equation, which are x = 3, and z = — 1; 
hence, the factors of (36) are « — 3, and x + 1. 

Now assume the equation 
(37) ih nad ae RS 

x’? — 2x — 3 x— 3 z+ 
where A and B are unknown quantities not containing x, and the 
19 


| 
| 


# = 
7 .- 


7 
1| 
4 | 
wy 
& 
7 
¢ 
i 
is 


198 DIFFERENTIAL CALCULUS, 


denominators of the assumed fractions are the factors of the deno- 
minator of the given fraction. Determine A and B by the method 
of indeterminate coefficients, which is done by reducing (37) to 
monomial terms, by clearing of fractions. This gives 

(38) az.= Ar + A + Ba —3 B, 

According to the method of indeterminate coefficients, the coeffi- 
cients of the like powers of 2 in (38) are equal. Hence, we have, 
(39) a=A+B, and o= A—BB, 

Solve these two equations for A and B, and we have, 


A=, and B=“, 


these values put into (37), we have, 
(40) Pe. ee ee el eee 
x? — 22 — 3 4 (x — 3) 4(x + 1) 
and the fraction is divided into two partial fractions. 
This procedure supposes the highest power of 2 in the denom- 


inator to exceed at least by unity, the highest power of 2 in the 


3 
numerator. If we had such a fraction as ; “i rc by dividing the 
eo — 


numerator by the denominator this may be put in the form, 
r+ ats 

x? — 5? 
where the last term only is a rational fraction, 

If some, or all of the factors of the denominator are equal, the 
process must be somewhat modified. Let the fraction be 

br + z* 
(e + 1) (e«—1)" 
where three of the factors of the denominator are equal. If we 
were to divide this as was done in (87), the division would fail, as 
will be found upon trial. But we may assume, 
(41) 
br + 2 abe Wwe. 2 C 4 D 

(@+1)(@—1f «$1 (¢«—1) (g2—1" 2—1’ 
where the second side shows the form of the partial fractions into 


RATIONAL FRACTIONS. 199 


which the given fraction can be resolved. The values of the con- 
stants A, B, C, D, in (41) are found as in (37), 

If there be equal sets of roots in the denominator of the proposed 

fraction, as in the form, 
ax 
(w — a) (w + bY 
we may divide it as follows. Assume 
(42) 
ax A B C D 

ee i a iF ; 
(x —a)? (x + 6) ~ (@—apy x—a wera a + bY r+b 
where the constants A, B, C, D, may be be found, as in (37). 

If some or all of the factors of the denominator be imaginary, 
the process requires another modification. If we have, 
(43) sa : ve 

(b + x) (x? + 2mzx + m + 2’) 
the denominator being equated to zero, shows the factors 

b+ a, Bink per eek BN an ya JCA 
where two of the factors are imaginary. And we may observe, in 
general, that the Theory of Equations shows that imaginary fac- 
tors always occur in pairs. Before dividing the form (48) into par- 
tial fractions, we may observe that the second factor of the denomi- 
nator may be put into the form, 
(a + m) + n’, 
and if x + m be put = z, this is simply 2? + n%, so that any 
form (43) may always be reduced to the simpler form, 
ax 
(bf a) (@ + my’ 

If the second factor in the denominator of (48) were, for exam- 
ple, 2? + x + 1, thisis tne same as (x + 1)? + 3, and if we 
assume z + 1 = 2z, it becomes 2? + 2, which agrees in form with 
the second factor of the denominator of (44). To divide (44) into 
partial fractions is therefore the same in effect as to divide (43). To 


divide (44), put 
ax A rs Ca D 


45 BAe me ee + ———— 
ae (5 + x) (a + nn’) b+e| B48 Pte 


(44) 


200 INTEGRAL CALCULUS, 


where A, C, D may be found, as in (37), by the method of indeter- 
minate coefficients, 
If we had several sets of imaginary factors in the denominator, 


the process would be as follows. Let 
2 


ax 
be the fraction. Assume the equation 
aw BAS Bz C D 


(46) (a? 72)? re (a? +2)? e+ ne (2 oh n)? 2 ne 

Here A, B, C, D may be determined by the method of indetermi- 
nate coefficients, as was done at (37). 

From the foregoing, we observe, that there are four classes of 
rational fractions, which may be divided into partial fractions. 

lst.— When the factors of the denominator are real and un- 
equal, as in (37). 

2d.— When the factors of the denominator are real, and some 
or all of them equal, as in (41), (42). 

3d.— When the denominator contains one set of imaginary fac- 
tors, as in (45). 

4th.— When the denominator contains more than one set of imag- 
inary factors, as in (46). 

As a generalisation of the first of these classes, suppose we had 


the fraction __°” __, where P.9;7,8, Soc, are the factors of the 
DOT 3~ ie 
denominator, and functions of 2. Then we assume 
: B C D 
(46a) Eder my Ee peas hae 
Rr Pies = Pp g r s 
Here A, B, C, D are to be determined by the method of indeter- 


minate coefficients. 
As a generalisation of the second class, suppose we had the frac- 


jel x ‘ 
tion pe where P is a function of x Assume this fraction equal to 
n 


the several partial fractions, as follows, 


te eh Ao, B C D M 
OD eee Ps Beg TPR es 


RATIONAL FRACTIONS. 201 


As a generalisation of the fourth class, put D.— a + n’, and 


2 
suppose we have the fraction a Assume this fraction equal to 


several partial fractions, as follows. 


ax? Az Ba Ma 
— > emery “ri. 4. Tica. +- qo 6 -_—_o . 
an ; a . : M ‘ 
y. f } , 
ss amas ———— e* © ——= @ 
pe a2 De" f D 


Determine A, B,...M, A’, B’,....M’, by the method of inde- 
terminate coefficients, and the second side of (46c) exhibits the par- 
tial fractions into which the first side may be divided. 

We will treat of the integration of each of the four classes of 
fractions now discussed. 

Case 1. 

Integrate a rational fraction when the factors of the denominator 
are real and unequal. 

Divide the rational fraction into its partial fractions, and integrate 
each partial fraction separately. 

Take the fraction 

ax dx 
sane Foe a 

Omitting the da, this is divided into its partial fractions at (40). 
Hence multiply (40) by da, and integrating each term of the second 
side, we have for the integral required, 

(a7) a ff ae aera 4 log.(e7—8) + © log.(e + 1) +. 

Ex. 2. Take the fraction 


ax dx, 
Here the factors of the denominator are «z, x—3, 2 +1. 
Then we may assume 


: A B C 

49 9° nas a Sige: 

So x? — 22°— 3x 7 sewer x+ il 
Multiply (49) by da, and integrating each term of the second 


side, we have the integral of (48), viz. 
i= 


202 INTEGRAL CALCULUS. 
(50) y = A log.x + B log.(x — 8) + Clog(z + 1) + C, 
where A, B, C may be determined from (49), as in (37). 

Ex, 3. Integrate the equations, 


adx Qa 3) da 
dy = — 2? dy = 4 shia 
xr’ —a xv + 2? — Q7 

Case 2. 


Integrate a rational fraction when the factors of the denominator 
are real, and some or all of them equal. 

Divide the rational fraction into its partial fractions, and integrate 
each fraction separately. 

Thus if we multiply (41) by da, the first side is a rational frac- 
tion, whose integral is the sum of the integrals of the fractions on 
the second side, each of which is obviously integrable, the first and 
last term being logarithms, and the second and third coming under 
Rule V, In like manner, if we multiply both sides of (42) by da, 
each fraction on the second side is integrable by Rule V. or VI. 
Their sum is the integral of the first side. 

Case 3. 

Integrate a rational fraction when one set of imaginary factors 
enters the denominator. 

Divide the fraction into its partial fractions, and integrate each 
partial fraction separately. 

Thus if we multiply (45) by da, the first side is a rational frac- 
tion, whose integral is the sum of the integrals on the second side. 
The two first terms on the second side are integrated by Rule VI., 
and the third term by (28). Hence, 

xdz 
EELS NE NET SH NG i 
(51) */, (0 + &) (x + #) me alge) 
a C log.(a? + n*) + D tan.) + C, 

2 n n 

Ex.—lIntegrate the equations, 


pei (a + bz) dz, 


2—] 


dx 


- dy 4. 
¥ e+e 


Case 4, 
Integrate a rational fraction when several sets of imaginary fac- 
tors enter the denominator. 


IRRATIONAL FRACTIONS, 203 


Divide the fraction into its partial fractions, and integrate each 


partial fraction separately. 
Thus if we multiply (46) by dz, the integral of each term on the 


second side is readily effected, except the third term. The first be- 
longs to Rule V., the second to Rule VI., and the fourth to (28). 


The third term, viz : 


(52) 


C dz 
(x? 2 nm)? 
may be reduced to depend for its integral upon the integral of the 


form mands , by diminishing the exponent of the denominator 
a +n 
by unity. But as a simple formula for reducing (52) will be given 
in a more general process hereafter, we will not here give the inte- 
gral of the form (52). 
Thus by dividing a rational fraction into its partial fractions, we 
may integrate an extensive class of differentials by the simple rules 


given in the first Chapter. 


CHAPTER III. 
INTEGRATION OF IRRATIONAL FRACTIONS. 


An irrational fraction is one that contains the variable under a 
fractional exponent. 

If the fraction consists of monomials only, it may be readily 
rationalised, and will then be brought under some of the processes 
of rational fractions. ‘Take as an example, 

1 
(53) ity toa Beara) 2%, 

2? — 2 : 

Assume 2 equal to z raised to such a power that the several frac- 
tional powers of x in (53) will be expressed in z with integer ex- 
ponents, This may always be done by putting x equal to z raised 
to a power given by the common denominator of the fractional ex- 


ponents, in the proposed term to be integrated. In (53) put 


1 
3 


INTEGRAL CALCULUS. 


, 1 
(54) . =i x? = 2, BSG and 
1 
xt = 2, also dx. = Aez de 
Put the values (54) of x in terms of z, into (53), and it becomes, 


_ (2° — m2") 12dz 
27 — 1 
In which, after dividing the numerator by the denominator, the 
integral will finally depend upon a rational fraction whose denomi- 
nator is 27 — 1. 
If we have to integrate an irrational fraction of the form, 
(56) SE aN 
(a + bx + c2*)* 
we must first rationalise the denominator, which may be done by 
some of the processes pointed out in the Diophantine Analysis. The 
procedure to rationalise such an example as (56), may be represented 
generally as follows. 


(99) dy 


Let us designate by 9(./x ), any combination of x under the 
radical. If then we have to integrate the fraction 


(57) i 
o(Va) 

we may assume 

(58) Q(/a) = F (2,2), 


where F (a,z) denotes such a combination of x,z and constant quan- 
tities as will, by the Diophantine Analysis, render the first side 
of (58) rational. Solve (58) for 2, and we may represent the result by 


(59) LIS 7 Be 

Differentiate this, and we have, 
(60) Ei wy aed. 

Substitute (59) into (58) second side, and we get 
(61) o( Jax ) = F (xz,z). 

Put (61) and (60) into (57), and it becomes, 

' dz 

62 ie Neaiag 
(0%) F (xz,2) 


which is rational, and may be integrated by some of the preceding 
methods. 


IRRATIONAL FRACTIONS. 205 


For the various modes of rationalising any given quantity, the 
student is referred to the Diophantine Analysis, whose principles 
and processes are explained in most books on Algebra. 

As a particular example of (57) take 

dx 
(a) ————___.. 
(a + Oe") 
In this case (58) becomes, 


(d) (a + b%x?)2 = bz. 2, 
Solve this for x, and we get 


a — 2 
Cc ie uo 
(¢) 2bz 


The differential of this is 
(d) eS i cadenay 
262? 
Put (c) into second side of (6), and we have, 
(e) (a + b2x2)2 = = a 
Put (d) and (e) into (a), and it becomes, 


b z 


whose integral is 
] . 
(g) Wie log.z +. C. 


Restoring in (g) the value of z from (5) we have, for the integra! 
of (a), 


(63) f ee 7 of log. (/a + ba? —bx) + C, 
(a - bx?) b 
which might be put into a more simple form. 

Before proceeding to rationalise (57), we may frequently put it 
into a more convenient form, by dividing the numerator and denom- 
inator by some of the constants that enter into the radical. Thus to 
rationalise (56), we might first divide the numerator and denomina- 


tor by Jc, or by /a, which would render the denominator readily 


rational. For, dividing by /c, and putting 
P 


= 


2 


eat a ee Re Es SSE 


r 
> 
‘ 

4 

7 


1) 
ie f 


ree 


~ ee 


= 


Se eh ae 


206 INTEGRAL CALCULUS. 


c =m, and Relays 4) 7 
, ¢ c 
the denominator of (56) is (m + ne + 2°)2, which is rationalised 
by putting it = z 4 z, 
If the denominator of (56) be divided by /a, then by putting 
b c 
- = MM, and —- =”, 
a a 
the denominator of (56) becomes (1 + ma + n22)?, which is 
rationalised by putting it = 1 + az. 
This last is the most convenient form, if c in (56) be negative. 
The student may integrate (56) for c positive and for ¢ negative. 
The same process would obviously apply if the denominator were 
of the form Fx.9./x, for then the irrational part might be ration- 
alised, and the rational part connected with it in terms of the new 
variable. Take for example, the equation, 
(h) dy = oc ED 
(1 + 2’) (1 — 2’)* 
The irrational part of the denominator is rationalised by putting 
ai— a2)? = 1 — zx, and equation (h) becomes, 
2 (2° -- 1) dz 
zat 627 + ] 
which is a rational fraction in Case 4. 
The denominator of (h) may also be rationalised by assuming 
(2) (l—#)? = (1—2)z, 
which leads to a more convenient form than (). 


CHAPTER IV. 
The object of this chapter is to integrate the Binomial Form. 


(64) ade (a + ba”). 
Here m and n may be taken as whole numbers. For if they 
were fractional, we could, as in (53), by putting a equal to z, raised 


INTEGRATION OF BINOMIALS. 207 


to a power denoted by the common denominator of m and n, change 
(64) into a similar expression where the exponents corresponding to 
m and n would be integers. 

We may also regard in (64) as positive ; for if it were negative, 
we could, by putting 2” = z”, change (64) to a similar expression, 
in which n would be positive. Nor need we have 2 in more than 
one term of the binomial (64). For if we had such a form as 
z”™—'dx (ax’ + bx"), this may be written a? + ™—'dz (a + ba”-*)?, 
which is the same in form as (64). If then we can integrate (64) 
when p is either whole or fractional, positive or negative, any of the 
other forms reducible to the form (64) may be integrated. 

To integrate (64), put 


ba” — ; ee ey Cae ay 2 oe EM (-)y 
(a) a+bz Z 7 ( ; ) x } 


and 


™ 


(0) aide = 1 () os de. 
n b 
The values from (a) and (6) put into (64), it becomes, 
— 4 
(65) oy (255°) Pdr, 
n b 


{tf now — be a whole number, or zero, the binomial of (65) is 
n 


rational, and (65) may be integrated by some of the preceding 
methods, 

Hence the binomial (64) may be integrated if the exponent of 
the variable without the vinculum increased by unity is divisible by 
the exponent within. 

This is the first condition of integrability. 

Take as an example, 

(c) dy = x°dx (a + bz*)* 

Here m = b, n = 2, p=1. The first condition of integra- 

bility is fulfilled, and (65) becomes, 


35 b* (z eed a)? z dz, 


a aie 
Snap ~Sr 


a 


ws 


208 INTEGRAL CALCULUS. 


where the binomial may be squared, and z*dz multiplied into it, and 
each term integrated separately. 

If the first condition of integrability be not fulfilled, we can get 
another condition, as follows. 

Divide (64) by x?”, and then multiplying by x”, it becomes, 
(66) xm ten—ide (aa + b)?. 

Put z = aa” + b, and by means of this, eliminate x and dz 
from (66), and it becomes, 


(66a) dh dasipy ( 


l— °) et. Ree 
nr 
a 

This is a rational binomial if “ + p be a whole number. 

n 

Hence the binomial may be integrated if the exponent of the vari- 
able, without the parenthesis increased by unity and divided by 
the exponent within, and this quotient increased by the exponent of 
the parenthesis is a whole number. This is the second condition of 
integrability. 

Kx.—Integrate the expression, 


(e) x (a + a)-3 da, 
Here m — 1 = — 2, n= 3, p= — 4, and the second con- 
dition of integrability is fulfilled, and putting 
(f) PL * eg fo (e) becomes 
(g) - = Sor ETE Aci SO Met 
z8 3a7z8 3a7z3 


These fractions are immediately integrable by Rule I. 


CHAPTER V. 
INTEGRATION BY PARTS. 


By the Differential Calculus, we have, 
(687) d.uv = udv + v du, 
From which, by integrating, we get 
(68) fu dv = vu rapt is du. 


INTEGRATION BY PARTS. 209 


This is called the formula for integration by parts, and is of ex- 
tensive application. By it any integral of the form fu dv is re- 
duced to depend upon another of the form fi v du. The use of the 


formula will be best shown by a few examples. 
Ex. 1. Integrate the expression, 


(a) xdx, log.x. 
Divide this into the two parts, xdx and log.z. Put 
(b) u = log.a, and de = «dz. 


Then differentiating the first, and integrating the second of (6), 
we have, 


(c) dy) = da and oo — i? 
x 2 
Put the values (6) and (ce) into (68), and we get 
d zdz.log.2 = a log.2 — aoe 
(d) frde. loge = log : 


Here the last term is immediately integrable. Integrating i., we 
have, for the integral of (a), 


\ dz, | ej ee 
Sg — ° —_—_—— ° 
(e) fe x, log.x 5 og.x : a 
Ex, 2. Integrate the expression, 
(f) sin.'z. dz. Put 
(g) Ui =a Gin, and do = dz. 


Differentiate the first, integrate the second of (g), and substituting 
into (68), we have, 
(h) . [sine ag = sin. 2 mat —zx*)-* x dx, 
where the last term is immediately integrable by Rule V. Integra- 
ting it, we have the integral of (h). 
Ex. 3. Integrate the expression, 
(k) tan.—'z.dz, 
Put tan.—'x = u, and dx = db, and (68) gives us, 
(1) /f tan a.dx = «x tan.’x —1} log(1 + 2?) + C. 
Ex. 4.. Integrate the expressions, 
cos,~'#.dz, versin.—'zx.dz, sin.'mz.da, tan.—'ma.dz. 
By means of the formula (68), for integration by parts, we are 


able to integrate many transcendental quantities. 
20 


210 INTEGRAL CALCULUS, 


Before proceeding further with these, we will show the application 
of formula (68), to the reduction of differential binomials. In the 
form, 


(69) x"—dx (a + bx”) . Put 


(m1) dp ards) and u=(a+ ba”). 

Integrate the first of (m), differentiate the second, and substitute 
into the formula (68), and we have, [writing for brevity < for 
a + bz"), 

(70) S x"—dx ge x” As ee pnb S aide gr, 
m m 

Here the integral of (69) depends upon another with different ex- 
ponents. This is one formula. 

The binomial (69) is obviously the same as 


71) an ada (a = ba”) h 


Put dv = (a + bx”) x”—da, and w= 2", Integrate the 

first, differentiate the second, and substituting into (68), we have, 
Hpbi 4. m—n 

(72 eda gy? ida ott1, 
Webi Ris nb (p+1) maeR ayy 

Here the integral of (69) depends upon another with different ex: 
ponents. This is another formula. 

—I 

Since (a + bx”) = (a + bx”) (a + bz") = azP"+ baz? 
(69) becomes, 

73) S 2” "dz.2? = a J pet rk ema ON 7 a eda gel 
which is another formula, where the integral of (69) depends upon 
two other integrals. 

Equate (73) and (70), and we get 
m _p > 

74 gibt get a Bee ig Ma x” "dr 2? , 

Ue) J b (m+ pn) b pang 

If in (74) we put p for p — 1, and m for m + n, it becomes, 
gta Pl __ a(m—n) 
b(m+pn) 8 (m-+pn) 
which is another formula. 


m—n—1 
yi z dx 2”, 


(75) 4 ae 2? dz = 


INTEGRATION BY PARTs. 211 


If we substitute for the last term of (73), its value in (74), we get, 


re a pna w 
(76) fur aedx so! ee Ma f ada . 
m+ pn m+ pn 


another formula. 
If in (74) we put p for p — 1, and solve it for the last term, 


we get, 

(77) fan—'da P= ee — eCocte ist. ove) funder’, 
; ma ma 

another formula, which is useful when m — 1 is negative. 

Solve (76) for the last term, put p for p — 1, and we have, 
x eget! | m+n-+pn fur dav, 

an (p+1) an (p+1) 
another formula, which may be used when p is negative. 

By the application of one or other of these formulas, the integra! 
of the differential binomial (69) may generally be reduced to depend 
upon an integral that may be immediately obtained. As an example 
of the application, take the following expression, 

(a) (a + b?x°)* dx. 

Here it may be desirable to diminish the exponent of the paren- 

thesis. For this purpose, compare (a) with formula (76), where 


(78) fa™—"dx 2? = 


b= m—1=0, or m=1, n=2, p = i, and we have, 
(5) S (a+ bx?) dx —— it (a-+B'x") oe iN p dx ' 


The last term of (b) is integrated at (63). Supplying the integral 
from (63), we have, for the integral of (a), 


(79) 
vi (a + bea) de=— (a+ bx’) 5 log.(./a+ bx? — br) + C 


As another example, take the following expression, 


(c) (R? — 2?)*de. 
Apply to this the formula (76), where a = R’, b= —I1,n=2, 
m—1=o0,orm = 1,p =}, and we have, 


(d) uJ (R? — we) de — = (R? — 2%)" Wy R [ da 
. a. 


y (Re— ay 


v| 8 


212 INTEGRAL CALCULUS. 


The last term of (d) is integrated at (21), viz: 
dx 


(R°2— at)? 

This put into (d), we have, 

; & SRE enh . 

| (80) f (R'—2")'de = > (R’ — 2°) ie sin i + C. 
As another example, take the expression, 


| (f) a’dx 


(€) 


= in Ci 
sl B + 


(2ra—a?)? 
3 _ 
This may be put in the form? (27 — x) 2d, 
As it would evidently serve no purpose, in such an example, to 


vary the exponent of the parenthesis by unity, let us diminish the 
exponent of the variable without the parenthesis, For this purpose, 


apply formula (75) where a = 27, 6 — — la=1, p=—i, 
andm—1 — 3, or m = 4, and (75) becomes 
(g) 2 : 

S vdx Qr—2)-t = — > (Qr—x) 42" f ahdx (2r—x)-t 


Apply (75) again to the last term of (¢), Where m — 1 = 1, or 
m == 3, and the other values remain as before, and (75) becomes, 


(A) of, a'dx (2r — x)-* =— am (27 — a)” + 7 dx “ 
(2ra—zx*)* 
The last term of (h) is integrated at (34), viz: 
(k) f Be = versin.—  , 
(27x — 2°) , 


Put the value of (k) into (h), and then the value of (h) into (¢), 
and we have, 


a’dx 


(2ra — x?)* 


x UP an a 
= —— lL (2raa? =(- 2ra—zx? 
y (2re—at)’ + "(ere 
(81) 
+ r versin.—! =) + C, 
r 
An extensive class of differentials may be integrated by the appli- 


cation of the formulas (70) to (78). ‘The same could be integrated by 
the direct application of the formula (68). 


TRANSCENDENTAL FUNCTIONS, 213 


CHAPTER VI. 
INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 


We have already seen, from (68a) to (68e), how the integration 
of logarithmic functions may be effected by integrating by parts. 
We have also shown the application of the same formula to the in- 
tegration of inverse circular functions, (68) to (68h). It remains to 
show how the same process may be applied to trigonometrical lines. 

Since by Differential Calculus, d. sin.a = cos.ada, we have, 
(82) a f cos.ade SSPE 


m sin.”—'x cos.rda, we have, 

slat! 2 

—__ ee 64 
m-+1 

In like manner, we get for the integral of cos."a sin.ada, 

cos." ) x 


and since d, sin.”"2 = 


(83) *, fsin.m2. cos.rdxz = 


(84) { cos.*x. sinedz = — up Hi 
nd n+l 
If we had to integrate the form, 
(85) sin,”2 cos."2dzx, 
we may assume 
sin.™y — u, and cos."x sin.xdx = dv. 


Differentiate the first, integrate the second, and substituting into 
formula (68), we have, 
Sige a COs tw 
n+l 


* sin,” 2 cos.” ada. 


efi ‘sin.”2 cos."adxz = — 


(86) m—|] 
| n+l 
Since by Trigonometry we have, 
cos."**x = cos."x (1 — sin), 
this value put into (86), it becomes, 
S sin.mx cos."x4dzx = — sin.” 'e cos," Fix 
m+n 


(87) ae sie 

m+n 

Here the integral is made to depend upon another where the ex- 
20 * 


Jf sine cos.”"4dx. 


214 INTEGRAL CALCULUS, 


ponent of sin.”z is diminished by 2. By the successive application 

of this formula, we can, when m is odd, reduce any differential of 

the proposed form to the form (84), and obtain its complete integral. 
Ex.—lIntegrate the expression, 


Hi (a) sin®z cos.2rda. 

i Comparing this to (87), we have, m = 3, n= 2, and (87) 
becomes, 

| in.2 3 

(bd) » Ssin2x cos2xda — — Si == + E Scos2a.sin.xdc. 


Integrate the last term of (b), as in (84), and substituting into (0), 
we have, for the complete integral of (a), 


3 
(c) S sin2x cos."2dx = — ~ = (sin.tx aa =) + C, 


Ex, 2.—Integrate sin.®a cos.tada. 


If m be even and n odd, we could, by taking as parts of (85), 
ae Riga I tt 7 and = sin."x cos.cdx = dv, 
and proceeding as before, obtain the formula, 
COs ait 


fh sin.”2 cos."ada = 
m+ n 


(88) omit 


mtn 
When 2 is odd, (88) reduces the form (85) to depend on (838). 
If n be zero, (85) becomes sin.”rda, and the form for integrating 
is found by putting n zero in (87) which gives, 


sin.”—"'z cos.z. ,° m—l1 7. 
(89) iii sin.”"rdx = — aie ri Bley ba 
m m 


iG cos." sin.”xd2. 


By the successive application of this formula, the integral will, 
when m is even, finally depend upon dx, and when m is odd, upon 


sin,edx, and thus the complete integral be obtained. 
If m be zero, (88) becomes, 

“Ilr sing . n— 

(90) SJ cos."ada = a + ; 

By the successive application of this formula, the integral will, 

when n is even, finally depend upon / de, and when n is odd, upon 


he cos.xdz. 


1 
ii cos.” axdzx. 


TRANSCENDENTAL FUNCTIONS. 215 


Ex.—Integrate cos.‘rdz. 
Employ formula (90) where n = 4. 


3, i 
cos.°a4 sin.x ae 
(d) a cos..ada = si + 3 J cos."rdx. 


Apply the same formula to the last term of (d), where n = 2, 


Cos.2 sin,z COS.2@ SiIn.a 
(e) S cos2xdx oes" eee. aes J de Pe adhe at ol 


t 
2 2 = 

Put this value of (e) into (d), and we have the entire integral. 

Ex.—Integrate cos.3rdx, sin.‘xdx, sin3rdz. 

If m be negative, the form (85) can be reduced by formula (88), 
to depend upon the integral of sin.-"2 cos.rdzx, when n is odd, and 
upon sin,-"xda2, when x iseven. The first of these is immediately 
integrable by (83). 

[t remains to integrate the form, 

(91) sin.—"rda. 

Solve (89) for the last term of the second side, and putting m for 
m— 2, we have, 

(92) f sin. xdx Sere age m+? f sin-xde., 
m+ m+] 

If m be negative, formula (92) will serve to reduce (91) to depend 
upon fat, when m is even, and upon 


09 f =, 
SIn.@& 


when m is odd. 
To integrate the form (93), multiply its numerator and denomina- 
tor by sin.2, and it may be put in the form, 


(94) afi sin.xdax 
1 —cos.*x 
Put == cosa -. —dz=sinadz, and 2 = cos.2z. 


These values put into (94), it becomes, 


z — dz 
95 f ik 
( ) 1 2 


This integrated as a rational fraction, we have, 


— dz Ls 
(96 SS =F oe (+) + 
) eye ON ee 4% 


216 INTEGRAL CALCULUS. 


or restoring the value of z, 
(97) aie xa ree) BLL, 
sin.x 1 + cos.2 

In a similar manner, if n were negative, (85) could, by formula 
(87), be reduced to depend upon the integral of cos.—"2 sin.adz, 
when m is odd, and upon cos.—"xdzx, when m is even, The first of 
these is immediately integrable by (84). It remains to integrate 
the form, 
(98) cos." rdz. 

Solve (90) for the last term of the second side, and putting n for 
n— 2, we get, 

Be yet sm EA COG Bene n+2 ae 

(99) J cos. cde” == aE + nar! xdx, 

By the application of this formula, (98) may be reduced to depend 
upon Ea dx when n is even, or upon 


(100) ia 
COS.L 


to la 


when 2 is odd, 
To integrate (100), multiply numerator and denominator by 
cos.x, and proceeding as with (98), we get, 


(101) dx v $ log (SE + sin. *) winger 
COS.& 1 — sin.z 
Eix.—Integrate 
sinZadx cos.2xdx dx dx 
cose” sinex COS.32 sin 


The preceding deductions of formule show the method of in- 
tegrating by parts. By means of the formula now deduced, many 
binomial and trigonometrical functions can be integrated. A very 
common practice, however, is to take the formula (68) for integra- 
ting by parts, and proceed, with its aid, to reduce the integral to 
some form readily integrable. In this way, transcendental functions 
may, in many complicated cases, be integrated. 

Transcendental functions may, in many cases, be changed to the 
form of algebraic functions, and treated by the methods already 
presented. ‘Thus to integrate 


INTEGRATION BY SERIES. 217 


| dx 
(102) ii ea cove eal 
a+ 6b cos.x 
we may put cosy7 = %, ~-- & = cos.—'z, and (102) is, by sub- 
stitution, changed to the form, 
—d 
(103) ‘ 


ee ee ee ne 

(a + bz) (1 — 2) 

which may be rationalised by the method of irrational fractions. Or 
if we assume 


(104) cos.2 = eer: 

and substitute into (102), it becomes, 
iad 2dz 

( 105) 


a+b+4+(a— b)z® 
which may be integrated by rational fractions, if 6 be greater than 
a, or by acircular arc, if 6 be less than a. 

But we will not pursue this subject further. The student who 
desires ample details of these matters, is referred to La Croix’s Cal- 
culus, volume II., where all the valuable labours of Analysts on the 
subject are collected. | 


CHAPTER VII. 
INTEGRATION BY SERIES. 


When none of the preceding methods enable us to integrate a pro- 
posed differential of one variable, the integral may be approximated 
to by a series. 

The Theory of Integration by Series may be readily understood. 
For let the expression to be integrated be represented by Qx.dr. 
Let this be divided into two parts, as g’a, and @’rdx, and let the 
part g'x be developed into a series by any of the methods of devel- 
opment applicable to it, [as by division, or by the Binomial Theorem, 
or by M‘Laurin’s Theorem, &c.], then multiply each term of this 
development by the other part, 9rd, and each term being then in- 
tegrated separately, we have the integral of the proposed 92.dz. 


218 INTEGRAL CALCULUS, 


As an example, let the expression to be integrated be, 


dx 
(a} as: 
; l+z 
To integrate this by series, divide it into the two parts : ; 
+2 
andda. The first part being developed by division, is 
1 
b =]— x? — x3 We, 
(5) force r+ a2—a4 && 
Multiply this by da, the second part, and we have, 
(c) sf dex = dx — adz + v'dxr — xdx. 
l+2 


Integrating this, we have, 
dx Si i at 
(d Se ae) &e. C. 
0 JS l+z2@z 2 3 4 as t 
This second side of (d) is an approximation to the integral. 
Ex. 2. Integrate the expression, 


1 — ex?)*d 7 
(105e) Sabi alot 
1 Neate 
This may be divided into the two parts, 
(1 — ex)’, and ae 
(1 — 2’) 
The first part developed into a series by the Binomial Theorem, is, 
a 2 py 4 a4 6.6 
( 1 — ey ey eee ex 8ex ee 
iy) ( ) ; us D4 wey c 
Multiply this series by the second part, pie es and the 


(1 — a?) 
second side becomes a series, of which each term may be integrated 
by the preceding methods. 
Ex. 3d. Integrate the expression, 
(g) (R?— x*)*dz. 
This may be divided into the two parts, (R? — z*)*, and dx. De- 
velope the first by the Binomial Theorem, and we have, 


3 Di et 
(h) “  (R? — 2’) = R— tes 34 R — Wwe. 


DIFFERENTIALS EXCEEDING THE FIRST ORDER. 219 


Multiply this by dz, the second part, and integrate, and we have, 


. 1 3 5 
k R? — 2°) de = Re —_=_. —_*___ — &&. +. 
age Age 23R  245R ss 
Ex, 4, Integrate by series, the expressions, 
(a? — 1)-td2, ad ew a A) hey 
zw+a (1 + a?)* a — a)é 


CHAPTER VIII. 


INTEGRATION OF DIFFERENTIALS EXCEEDING THE FIRST ORDER. 


The second differential coefficient, third differential coefficient, &c., 
are called differential coefficients of the higher orders. Suppose we 
have, 


3, 
(106) TY = me. 
dx* 
Multiply by da, and we have, 
3 
(107) vy | made. 
daz? 
Integrate and we have, 
? mx? 
108 Be Ate! SEONCEP °C. 
eid dx? 2 
Multiply again by dz, and integrate, and we have, 
(109) = = - Om 5.0". 
Multiply again by dz, and integrate, and we have, 
4 Ca? 
110 fetish ee rein" 
on) i 2.3.4 2 i a5 


This is the complete integral of (106), a constant being added at 
each integration. This example shows the process of integrating in 
such cases. 

4 5 

Integrate oe Mx”, and ay 

dx‘ dx* 


An integral of the second order is denoted by the form Wie, or by 


220 INTEGRAL CALCULUS. 
repeating the sign of integration, thus J/- (n the same way, the 
third integral of axda*, may be denoted by 
af ada’, Duby a Sf frde. 
We shall now give a number of Geometrical Applications of the 


Integral Calculus, which we will put into the form of Propositions, 
for the sake of convenience. 


PROPOSITION I, 


Determine the area of a plane curve. 
FIG. 77. 


| Let AP be the curve, whose equa- = P ans 
tion is 
(111) y = v2. 

Let (2,y) be the point P on the 
curve. ‘Then da may be represented 
by BC. Let the area be reckoned be- 
tween the curve and the axis of X. If 
ABP be the area, BCQP would repre- A. BC = 
sent the differential of the area, being an indefinitely small incre. 
ment of it. Now, by the figure, we have, ydx = area of the rect- 
angle BR. When BC or dz is indefinitely small, the rectangle BR 


‘becomes the differential area BCQP, that is, putting A for the area 


of ABP 

(112) dA =\ ydz, 
and integrating this, 

(113) A = f ydz. 


Such is the general form for the area contained between the curve 
and the axis AX. 

The integration of (112) was regarded, by the early writers on 
the Calculus, as the addition of the elemental areas BR, or rather 
BCQP. 

To apply (118) to any given curve (111), weeliminate one of the 
co-ordinates in (113) by means of the equation of the curve. Thus, 
substitute from (111) into (113) the value of y, and we have, 


OF TRE ARBITRARY CONSTANT. 221 


(114) A = fou.de, 
which being a function of a single variable, may be integrated. [See 


Problem F, post.] 
Ex.—Find the area of the common parabola. 


Here (111) becomes, 


(a) y = pa’. 
This value of y put into (113), we have, 
? 7 $ 3 ; 
(b) A=p fxde== prt +C, 


the area required. 
Before proceeding further with examples under this Proposition, 


we will solve the following Proposition. 
gw 


PROPOSITION II. 


Determine the constant added in integration. 

To explain the method of doing this, take the following exainple 
under the preceding Proposition. Find the area of the curve, 
(c) y = ax —b, 

Put this value of y into (1138), and integrate, and we have, 
(d) Als = SP beh C: 


This is the general expression for the area. 
FIG, 78, 


If now we wish the area PBD, i 
bounded on the abscissa by BD, we 
must obviously subject the integral 
(d) to such conditions as will express 
this area. | 

Let AB = m, and AD=n, [f A 
we estimate the area from B towards 


PD, it is obvious that when 2@ = m = AB, the area (d) is zero. 
Put then in (d), x = m, and it becomes, 

um 
(e) o= — bm + €, 


Q 21 


Se ee ae 


a “all 
a 


a 


INTEGRAL CALCULUS. 


This equation is sufficient to determine the constant C, and the 
value of it from (e) put into (d), would give us the area of PBD for 
any abscissa x. If a definite value m be given to AD, then if 
x = n, (d) becomes, 


(f) Area = 2 — bn + C, 
If (e) be subtracted from (f), we have, 
(g) Area, = > (n? — mm) — b (n — m). 


In this we have the area required, independently of any indeter- 
minate constant, 
If (e) be subtracted from (d), we have, 


(A) Area = = (x? — m’) — b (x — m), 


where the constant is also eliminated, and the base BD of the trian- 
gle becomes fixed by giving to x any value greater than m. 

Equation (d@) may be regarded as containing three unknown quan- 
tities, viz: #,C, and A. The conditions to which we subject the 
particular proposition, enables us to determine two of them, viz: x 
and A, and gives (e) an equation which makes known the value of 
the indeterminate C. 

This is called Integration between Limits. ‘The limits that give 
the determinate area (g) are x = m, anda =n. This is usually 
denoted by writing the limits in the form of a fraction after the sign 


n . ° ° * . . 
of integration, thus f m, the subtractive limit being in the denomi- 


nator. Thus the expression, (i ; (ax — x’) dz, means that the 
constant after integration is to be eliminated by taking the integral 
between the limits « = 0, andaz =n. Equation (h) is made by 
taking the integral between the limits a = m, and x = @. 

The constant may be zero, but Integration between Limits is the 
process substantially employed, in all cases, to determine the con- 
stant and define the integral, The object aimed at in each particu- 
lar proposition, must designate the limits, between which the integral 
is to be taken. 

We will add some examples illustrative of these two propositions. 


OF THE ARBITRARY CONSTANT, 2238 


Ex, 1. Determine the area of a common parabola between the 
limits 2 = 0, anda = @, 

This is done by taking equation (6), Proposition I. between these 
limits, For 2 = 0, the area is zero, and (6) becomes, 0 = C. 
Hence (5) is simply 


(k) A= pt 


by eliminating p*by means of equation (a). Hence the area of the 
common parabola reckoned from the vertex is two-thirds of the cir- 
cumscribing rectangle. 

Ex. 2. Determine the area of the circle. 
(1) y = (R’— a’, 
is the equation of the circle. This value of y put into (113), we 
have, 


(m) A= /(R? — 2°)'dz, 
an equation integrated at (80), where we have, 
x 3 R?) ig. x 
n ~ A> — (R?— 7? Bie ( 
() ou Lene: ot 


This between the limits 2 = 0, and x = R, gives the area of a 
quadrant of the circle. Estimating the area from the axis of Y, 
when x = 0, we have, A = 0, and C = 0, Hence C may be 
omitted in (n). For the other limit, if « = R, (n) becomes, 

(0) ta Sein“, 


ad 


But snl = 90° = > Hence (0) is 


R? 6 
(P) A = a a quadrant ; 
and four times this is, 
(7) circle = 7R?, 


which agrees with the result in Plane Geometry. 
Ex. 3.—Determine the area of the ellipse. 
The equation of the ellipse solved for y, is, 


oi ky ola 
(7) Y =e(G x’). 


224 INTEGRAL CALCULUS. 


This value of y put into mete) we have, 
(s) We 2 fe a) de. 


The part under the sign of integration is by (m) the area of a circle 
whose radius is a@. Hence if a circle be described on the major 
diameter of an ellipse, the area of this circle multiplied by the minor 
diameter, and divided by the major diameter, is the area of the 
ellipse. ‘ 

Ex. 4. Determine the area of the cycloid. » 

The differential equation of the cycloid is, [Appendix (22), Differ- 
ential Calculus], 


(2a) dx = yay 


(2ry — yf) 
This value of dz put into (113), we have, 


__ ydy 
(25) es 5 
(2ry — 9)" 
This is integrated at (81), from which we have, 
(2c) 
A=— = (2ry—y?)" ms a (—(2ry—-9") +7 versin.— vy +C. 
‘he 
Between the limits y = 0, y = 2r, this gives us the area of 
ADE, half of the cycloid. Wheny= 0, A = o, and C = 0. 


FIG. 79 


When y = 2r, (2c) be- ¥ 
comes, 


% ape Be 2) Soir. 
Q 2 
(2d) wee versin.'2, 8B 


But A 
VOrsiNe sy? c= LO0. == 7, as Wenn SWI Se B 
This put into (2d), and that equation doubled, we get 

(2e) area of cycloid = 3ar’, 


which shows the area of the cycloid to be three times the area of the 
generating circle. 
Bere eenine the area of the following curves :— 
ee Ee ieee 0a, y = asin.a, y = asin.‘z cosa. 
This last value of y put into (113), we have, 


RATIFICATION OF CURVES. 225 


(2f ) Ae af sin.“z cos.*rdz, 
whose integral is found by (88). 

Cor. 1.—If we take, fig. 79, BC = dy, and BF = 2, we would 
find, instead of (118), 
(115) dy Aa == zdy, 
where the area would be the part between the axis of y and the 
curve. If the equation of the curve be 


ae r= Vy; 
this value of # put into (115), and that equation integrated, we have, 
(117) A = f yy.dy. 


Cor, 2.—In determining the differential area (112), the axes were 
supposed rectangular, or BR, fig. 77, was a rectangle. If the axes 
be inclined at an angle 9, it is obvious that ydx must be multiplied 
by sin., to express the differential of the area. In this case (113) 
becomes, 

(118) Aye sin.g f ydx. 

As an example of the application of (118), find the area of the 

curve, 

(2g) xy = mm, 

which is the hyperbola referred to its asymptotes. Let o be the 
angle of inclination of the asymptotes. ‘The value of y from (2g), 
put into (118), and that equation integrated, we have, 

(2h) A = m sin.g log.x + C, 

which may be taken between any proposed limits. 


PROPOSITION III. 


Determine the length of a given curve. 
FIG. 80. 


are (x,y) be the co-ordinatesof P, * P 
Fig. 80. 

Let Q be indefinitely near to P, and 
we have, dz = PR, dy = QR. 

Let AP = zy os PQ = dz. 

Suppose the curve a polygon of an 
indefinite number of sides, one of 
which is PQ, then from the triangle 4 

: cs FM 8 
PQR, [the axes being rectangular], we have, 
ai* 


226 INTEGRAL CALCULUS. 


(119) dz? = dz* + dy® «. dz = (dz? + dy’). 

The integral of this gives the arc 2, viz: 
(120) z= f (de? + dy?) 

If the equation of the curve be 
(121) y = 9, 
we can, by means of this equation, express the second side of (120) 
| in terms of one variable, and integrate it. We may put (120) into 
H the form, 


(122) % =f(i a p*) de, where’ p= a or, 


(123) z =f(l + q’) dy, where —— 
y 


which are convenient in practice. 

Ex. 1. Determine the length of an arc of the parabola. 
(a) y = Amz, 
is the equation of the curve. Differentiate this, and we have, 
(b) 42) 39 wolgtotlgca.og 

This put into (123), we have, 

1 2 
‘4 = Am? *\ ay, 

(¢) = J (Am? + y') dy 


This is integrated at (79), by making a = 4m’, and b = 1, from 
which we have 


(d) 2= a (4m?+-y) — m log.( /4m + y? — y) + C. 


This taken between the limits y = 0, toy = y, will give the 
length of the curve for any ordinate y. For y = 0, the arc z = 0, 
and (d) becomes, 

o=—mlog2m+C, “ C= mlog.2m, 
which may be put into (d). 

Ex. 2. Determine the length of the cycloid. 

From the differential equation of the cycloid (2a), last Proposi- 
tion, we get, 

DOTA S ye) Ueatin 058 
(f) G5 lec cae 


RECTIFICATION OF CURVES. 227 


This value put into (1238), we get, 


£) Z= J2rf (2r — y)-*dy. 
This integrated by Rule V., we have, 
(A) 2=—2 /2r (2r — yy C, 


For half of the cycloid AD, [fig. 79], take (2) between the limits 
y = 0, y = 2r, when y = 9, z is zero, and (h) becomes, 


(k) o=—4r+C, 
When y = 2r, (h) becomes, 

(L) x = C, 
Subtract (/) from (Z), and we have, 

(m) @ = hay, 


The double of this, viz: 87, is the length of the whole arc of the 
cycloid. 

Subtract (k) from (h), and we have, 
(n) “ 3 = 4p —2 /2r (27 — y)*, 
which is the length of an arc DP, measured from the vertex D to 
any point P whose ordinate is y. ' 

Ex. 3, Determine the length of the ellipse, 

Take as the equation of the ellipse, 
() y= (2 —1) (0? —a’), 

The value of dy deduced from (0), in terms of 2, and put 
into (120), gives, after putting 2 = az’, 
(p) y= af (i= nib 

(1 — x'*)* 

which may be integrated in a series, as directed at (105e). 

Ex, 4. Determine the length of the semicubical parabola. 

y° = az’ is the equation. “Let the limits be x = 0, 2 = m., 

Ex. 5. Find the length of the curve, 


” 


° 2 
x? + y% = m?, between the limits 7 = 0, x = m. 


3 
Ans. 4—= -_ ™. 
5) 


228 INTEGRAL CALCULUS. 


PROPOSITION IV. 


Determine the area of a surface of revolution. 


If the curve AB revolve round the fixed axis AX, it generates a 
surface of revolution, For the purpose of applying the Integral 
Calculus, the surface of revolution may be regarded as generated by 
the perimeter of a circle which moves with its centre on the axis of 
revolution AX, has its plane perpendicular to that axis, and the 


curve AB in the plane of the paper for its directrix. 
FIG. 81. 
Let the equation of the directrix 


be, 

(124) Y = PU, 

and let (x,y) be the point P. Since 
y is the radius of the generating 
circle, 22y is the perimeter of the 
generating circle, 

Suppose the points’P and B in- 
definitely near to each other, then 
the half sum of the perimeters of 
the circles whose diameters are PE H 
and BH multiplied by PB, gives the area of the conic frustum gene- 
rated by the revolution of PB round AX. The area of this conic 
frustum is the elementary increment of the surface or differential of 
the surface. But since P and B are indefinitely near, the perimeter 
of PE may be taken as the half sum of the perimeters of the circles 
PE and BH. Hence calling the surface of revolution S, we have, 
(125) dS = 2ny x PB. 

Substitute in this for PB, its value dz in (119), we have, 

(126) dS = 2ny. (dx® + dy?)}, 
and integrating this, we have, 
(127) S'= 27 fy (da? + dy?)?. 

By means of (124), the equation of the directrix, the second side 
of (127) may be expressed in terms of a single variable, 


SURFACE OF REVOLUTION, 229 


Ex. 1. Determine the surface of the sphere. 
(a) y = (2rx — z”*)*, 
is the equation of the directrix AB in this case. From (a) we get, 


__ (r— 2)! 5 
b CA eee, OE 
(9) Y 2rx — x 
This value, and that of y in (a), put into (127), we have, 
(c) i ar [dx sqennn, tC, 
This between the limits a = 0, x = 2r, gives for the surface of 


the sphere, 
(d) S = 4xr° = four great circles. 
Ex. 2. Determine the surface generated by the revolution of a 
cvcloid about its base. 
"The differential equation of the cycloid is, [Appendix, Differential 
Calculus, (22), ] 
ydy 
(2ry—y?)* 
This value of dx put into (127), we have, 
(f) S = 29 y2rf (2r —y) *ydy. 
This integrated by formula (75) is, 


(ig) S=— Ee n f2ry (2r — y) — = arf 2r (2r—y) + C. 


(€) dx = 


This between the limits y = 0, y = 2r, gives for the area of half 
the surface, 


h Se ea 
(i) 2 


and double of this is the whole surface. 
Ex. 3. Determine the surface of the cone. 
Here y = az is the equation of the directrix. 
Ex. 4. Determine the surface of the ellipsoid of revolution. 
Ex. 5. Determine the surface of the paraboloid of revolution. 
Ex. 6, Determine the surface generated by the revolution of the 


° 


9 
curve, x3 + y3 = m3, round the axis of X, 
This between the limits x = 0, 2 = m, gives for the surface, 
d D : 
(ke) S = 22m’? = areas of twocircles of radius m. 


jis 


t 


: 
i 


hE Taek = 


230 INTEGRAL CALCULUS. _ 


Ex. 7. Determine the area generated by the revolution of a 
cycloid round its axis, [DC fig. 79.] 
Ans. S = 8nr? (1 — 4). 


PROPOSITION V. 


Determine the volume of a solid of revolution. 


The solid of revolution may be conceived to be generated by the 
area of a circle moving as in last Proposition. 
FIG. 82 


If (a,y) be the point P, and 
(128) Y = Ou, 
the equation of the directrix AB, 
then the area of the generating cir- 
cle is zy’. This multiplied by 
dx = OR, we have, xy’da, which 
is the volume of the cylinder whose 
base is the circle PE, and whose 
altitude is OR = dz, 

When the points P and B are in- 
definitely near, this cylinder be- ors 
comes the same as the solid EHBP, which is the differential of the 
volume. That is putting V for the volume, we have, 


(129) dV = xy'dx, 
This integrated, gives the volume, 
(130) Vide nf y'da. 


By means of (128), this may be expressed in terms of one varia: 
ble, and integrated. 


Ex. 1. Find the volume of the sphere. 
(a) Ty oe 
is the equation of the directrix. This put into (130) and integrated, 
we have, 


(b) Vea(r—Z)ic. 


VOLUME OF SOLID OF REVOLUTION. 231 


This between the limits 2 = 0, x = 2r, gives for the volume of 
the sphere, 
mies 
c Nast nT”. 
(0) ; 
Ex. 2. Find the volume of the ellipsoid of revolution, 


(4) y= 2 @—2'). 
7 


This put into (130), and integrated between the limits z = 0, 
x = a, give the volume of half the solid. 
Ex. 3. Find the volume of the paraboloid of revolution, 


(e) Y= Ni, 

is the directrix. This put into (130), and integrated, gives, 
vi mpx 

(f) SU 


which needs no correction, forC, V, and 2 are all zero together. 
Eliminate p between (e) and (f), and we have, 
(gyo- View ae equal half the circumscribing cylinder. 

Ex. 4. Find the volume of the cone. 

y = az, is the directrix. 

Ex. 5. Find the volume generated by a cycloid revolving round 
its base. 

Cor.—If the axis of y be the axis of revolution, instead of (130) 
we would obviously have, 
(131) Vix a f x'dy. 

Ex.—Find the volume generated by the revolution of a parabola 
round the tangent to its vertex. 

The axis of y is the axis of revolution, and the value of x from 
(e), put into (181), and that equation integrated, we have, 


(h) V= TY 


~ 


5 
Ex.—Find the volume generated by the revolution of the curve, 


9 ° 9 
r% + y3 = m3, first round the axis of y, second round the axis 
of x, between the limits y = 0, y = m, and x = 0, = m. 


Ans. v= 16 m. 
105 


a 
a 7 


<<a 


SOS FS Ss 


on ee ee 


SS oo ee FS Sea rk ores se 


a 


b 
a 
WA 
| 

; 


232 is INTEGRAL CALCULUS. 


PROPOSITION VI. 


Determine the area of a plane curve referred to polar 
co-ordinates. 

Let BE be the curve, AB the angular axis, AC the radius vec- 
tor = p, CAB = a, and CD perpendicular to AC. Then by Dif- 
ferential Calculus, [Proposition XXXIII., (169),] we have, when Ei 


and C are indefinitely near to each other, CD = pda, 
FIG. &3. 


Then the triangle ACD has 


2 
for its area, i , which 


may be taken to be the same 
as the area ACE, the elemental 
area or differential of the area A B 
CAB. Hence if A be the area, 
(132) dA w= be A 
and integrating this, we have, 
(133) = 2 f p'du. 
If the equatign of the curve be, 
(134) oes 
we can, by means of this, express (183) in terms of one variable, 
and then integrate it. 
Ex. 1. Determine the area of the curve, 


(a) p = Ga, between the limits p = m, and p = n. 
Differentiate (a), and we have, 
d 
b de = —, 
(0) a 
which put into (183), we have, 
1 1 
A ee 2S See 
(c) ‘ J Pap mae , 
and this between the limits proposed is, 
1 
d A => — ne ey m> e 
(4) aya ) 


POLAR AREAS. 223 


We might also have found this in terms of o, by putting into 
(133) the value of p in (a). ' 

Ex. 2.—Determine the area of the circloid. 
(e) p = a C0S.a + a, 


is the equation of the curve, [Differential Calculus, (136f ),}. 
FIG. 84. 


Let us express (133) in terms of p, For 

this purpose, differentiate (e), and we get, 
- dp 

(J) Sor iat @ SiN. 

Put into (e) for cos.w, its equal, 

(1 _— sin2w)*, 

and solving the result for sin.o, we get, 
(g) a sin.w = (2ap — 9°), 
which put into (f ), gives, 


dp 
(2ap—p*)* 
This value put into (183), we have, 
1 p-dp 
(k) irAey =f wea ER 
2 (2ap—p ) 


This is integrated at (81), from which we have, 


2 a ° 
QA = Ea p (2ap—p*) —F(-Ca—') +4 versin—'2) + C. 


a 


(h) de = 


This taken between the limits p = 0, and p = 2a, we have, 
3 9 
(m) ees 7% 7, 


which is the area above the angular axis AB. 
The double of (m) gives the whole area of the circloid, viz: 


(n) circloid = hit wa”. 
2 
Hence the area of the circloid is three semicircles of the direct- 
ing circle. 
If we substitute into (133) for p?, its value from (e), we have, 
(0) A =  f (a’cos."wd a + a’dw + 2a*cos.wdw), 


22 


234 INTEGRAL CALCULUS. 


where each term may be integrated separately, the first term under 
the vinculum being of the form (90). The integral of (0) between 
the limits, » = 0, and » = 180°, will, as before, give us (m). 

Ex. 3. Determine the area of the curves, 
A p = dy", o* = a’cos.2a, from w = 0 to # = 180°, 


PROPOSITION VII. 


Determine the length of a curve referred to polar co- 
ordinates. 
FIG. 8&5. 


Let BC be the curve, and z 
its length. Then EC repre- 
sents dz, and the elemental Cc 
triangle EDC gives, 

(135) EC? = CD? + DE’, 

which since DC = pda, and 4 

DE = dp, becomes, 

(136) dz = (p*da® + d,?)*. 

Integrating (136), we have, 

(137) « Zz = f (p’dw* — dp*)s. 

If the curve be represented by 
(138) p = Qa, 
by means of this equation we may express the second side of (137) 
in terms of a single variable, and then integrate between any 
assigned limits. 

Ex. 1, Determine the length of the curve, 

(a) p = dw. 

By means of (a) we may eliminate either p or w from (137). To 
eliminate w, differentiate (a), and substitute the value of dw into 
(137), and we have, 


(b) 2 =f (a + si)hdp, 


which is integrated at (79). 


DUE 


AREAS BY DOUBLE INTEGRATION, 235 


Ex. 2, Determine the length of the circloid. 
(c) p = @COS.w + a, 
is its equation. [Differential Calculus, Proposition XXIV., (136f).] 
Put the value of dw» from (hk), last Proposition, into (137), and 
we get, 
(d) . = J/2af (2a —p)*dp = —2 /2a (2a— pF + C. 
This between the limits p = 0, and p = 2a, gives for the length 
of BPA, fig. 84, half of the curve, 
(e) z == 4a; 
and the double of this, 8a, is the whole length, which is therefore 
equal to the perimeter of the square circumscribing the directing 
circle. 
Ex. 3. Determine the length of the curve, 


(fF) pv = a, 


PROPOSITION VIII. 


Determine the area of a plane curve by double integration. 
FIG. 86, 


Let BC be the curve. 

Let (x,y) be the point N. 
Let the point S be indefi- 
nitely near to N. Then 
i> DE, dy = NK, 

Make LM = NK = dy, 
and it is obvious that 
(139) dzdy = FGSK. 

Now if A = area of the 
curve, -- dA = DESN, 
as was shown in Proposition I. 

But FGSK may be regarded as the differential of DESN, and is 
consequently the differential of a differential, or the second differen- 
tial of the area, that is, 

(140) PA = dzdy, 

If this be integrated twice, we have, 


236 INTEGRAL CALCULUS. 


(141) A = ff dxdy. 
By inspecting the figure, it is evident that FG or dx may remain 
constant while dy or KF is repeated along ND. Hence one of the 


integrations of (140) may be made for dz constant, and dy varia- 
ble. This gives, 


(142) dA = ydz. 
If now the equation of the curve be, 
(143) Yi SA; 


the constant added to (142) may be eliminated by taking (142) be- 
tween the limits y = 0, y = gx, by which (142) becomes, 


(144) dA = oxdz. 
Equation (144) integrated, gives, 
(145) A = f[ pxdx — i rae OS 


denoting the integral of grdzx by g'x. If the constant in (145) be 
eliminated by taking that integral between the limits 2 — Deo Ate, 
we have the whole area, ACB. 

The integration of (140) for dx constant and dy variable, may 
be regarded as the addition of the elemental rectangles KG, in the 
differential area DS. 

Again, by inspecting the figure, it is evident that ML or dy may 
remain constant while FG or dz is repeated along LR. Hence one 
of the integrations of (140) may be made for dy constant and dx 
variable. ‘This gives, 


(146) dA = «dy. 
If now the equation of the curve be, 
(147) T= VW; 


the constant added in (146) may be eliminated. by taking it between 
the limits 2 = 0, x = yy, by which (146) becomes, 


(148) dA = Aydy. 
This integrated again, gives, 
(149) A= frydy = vy + ©, 


where 4’y is put for the integral of .ydy. If the constant be deter- 
mined by taking this integral between the limits y = 0, and y= AB, 
we get the area between the curve and the axis of y. The integra- 
tion of (140) for dz variable and dy constant, may be regarded as 


VOLUME OF A SOLID. 237 


the addition of the elemental rectangles KG in the differential area 
MR. The integral (145) is obviously the same in signification as 
(113), and the integral (149) the same as (115). 

The object of this Proposition has been to familiarise the student 
with the Geometry of Double Integrals, as preparatory to the suc- 
ceeding investigations. 


PROPOSITION IX. 


Determine the volume of any regular solid. 

Proposition V. enables us to find the volume of a Solid of Revo- 
lution, but the object of the present Proposition is to deduce a 
method applicable to any given solid. 


Let the equation of 
the surface LCB be 
(150) «= 9 (a,y). 

Let CB, the inter- 
section of the surface 
with the plane xy, be 
represented by the 
equation, 

(151) pate 

Let (x,y,z) be the 
point Q on the sur- 
face. ‘Then taking the 
point S_ indefinitely 
near to Q, and passing 
through each of the points Q and S two parallel planes, we have the 
figure, in which, 


AF =<, Bikiveirday 
(A) BR aby! and RH = dy, 
RQ = z, RN’ ='QL' = dz. 


If V be the volume comprised between the co-ordinate planes, 
and the parallel planes D'FU and BET, then, 
ID 


R PP Ai 


. 
1 
) 


| 
| 


238 INTEGRAL CALCULUS, 


(152) dV = solid D'QRKGE,, 
and the part QPHKG' is the differential of the solid D'QRKGE’, or, 
(153) a’V = solid QPHKG’, . 
and the solid HKOM is the differential of the solid QPHKG’, or, 
(154) d’V = solid HKOM., 
But it is obvious that by notation (A), 
(155) solid HKOM = dadydz. Hence, 
(156) @V = dudydz. 


The third integral of this is, 

(157) V = fff dudydz. 

By inspecting the figure, it is obvious that dx and dy may remain 
constant, while dz is repeated along the line QR. That is, (156) 
may be integrated for z variable, and dx and dy constant. Inte- 
grating (156) once on this hypothesis, it becomes, 

(158) PV = dxdyz, 

or, which is the same, 

(159) inde =e 
dady 

This may be corrected by taking it between the limits z = 0, 
z= 9 (x,y), [see (150),] which gives, 

a? 


160 aT SN A 
( ) dady P (x y) 
which is the whole line QR. 
By the figure, it is obvious that dx may be constant, while dy is ; 


repeated along the line FU. Hence (160) may be integrated for dx 
constant and dy variable. Multiply (160) by dy, and integrating 
for y variable, we have, 
dV | 

(161) enw di o(2,y) dy = g'(a,y) + C, 
where 9'(2,y) is put for the integral of o(a,y) dy. 

This may be corrected by being taken between the limits y = 0, 
y = yr, [see (151),] which changes (161) to 


(162 S" = E(2,12). 
(162) oF Q'(X,px) 


Multiply this by da, and integrate, and we have. 


VOLUME OF A SOLID. 239 


(163) . V = [ 9'(«,y2) de = oe 
which may be corrected by taking it between the limits x — o, 
«x = AB, or any other assigned limits. 


It is obvious that by taking (158) between the limits z = 0, 
% = 9(«,y), we get the solid whose lower base is the parallelogram 
HK, and upper base the surface QS. It is also obvious that by in- 
tegrating (160) for y, asthe only variable, we get the area of the 
parallel section D'FU, which area is the second side of (162). And 
ater multiplying (162) by dz, in order to integrate, we see that the 
solid is the parallel area D’FU multiplied by dx, or FG. Hence we 
may regard the area D'FU as the generator of the solid, that area 
moving parallel to itself, and varying so as always to be a section 
of the surface, 

Instead of integrating (156) for z, as the only variable, we might 
have integrated it for x or y, as the only variable. In (160) also, 
we might have integrated first for # instead of y. 

In practice, it is usual to begin with (159), or rather with (160), 
its equal. 

llustrative of this theory, take the following : 

Example 1. <A paraboloid of revolution is cut into a wedge by 
two planes, whose line of intersection passes through the vertex, and 
is perpendicular to the axis of revolution, the base of the paraboloid 
is the back of the wedge, find its solid content. 


FIG. 88. 


Let AX be the axis 
of revolution. 

Let the planes that 
cut the paraboloid be 
perpendicular to XY. 
AZ is their line of in- 
tersection, and CABD 
is one-fourth of the 
wedge. Let the equa- 
tion of AB, the trace 
of one of the cutting planes on the plane XY, be 


=a eo ype eres 
ao a Lp — 2 =z ——————— 


re 


EET SE 


=: ~ aS 


240 INTEGRAL CALCULUS. 


(a) y¥ = ar, 
and the equation of the paraboloid, 
(6) z= (pr — y’). 
This value of z, put into (159), we have, 
av avi 
(c) dady = (pr—y’). 
Multiply by dy, and integrating for a constant, we have, 


V 
(d) u 


— pr —y F . 
a J (pe — yyy 

The second side of (d), for x constant, is integrated at (80), from 
which we have, 


dV sd 2? Pe a a 
€ — oo tC— ee . 
(¢) Pat Merah OLA wort bo Sng ha ee 
This taken between the limits y = 0, y = az, we have, 
: dV _ ax onayt , PE. _; © 
(tT ) ree Sone ax") Peeouca aN) = 


Multiply by dz, and integrating, we have, 
(F)quve = eeh(p — ax) 'x?dx + L fx sin-—a [=. da, 
2 2 \ Dp 


the first term of which fulfils the second condition of integra- 
bility, (66). The second term may be integrated by parts. 
For the whole quarter of the wedge, (g) must be taken between 
the limits 7 = 0, « = dim 
a2 
Ex. 2. A given cylinder is cut by two planes into a wedge, 
having a diameter of one end of the cylinder for its edge, and the 
area of the other end for its back, find the volume of the wedge. 
Ans. V = MR? (x— 4), 
where M = axis of the cylinder, and R = radius of the base. 
Since ~MR? is the volume of the cylinder, 4MR?’ is the volume 
of the parts cut off in forming the wedge. The volume of these 
parts is therefore a geometrical cube. 


AREA OF A SURFACE. 241 


PROPOSITION X., 


Determine the area of any regular surface. 
FIG, 8. 


Let CLB be the sur- 
face, and its equation 
(164) z= (zy). 

Let (a,y,z) be the 
point Q, and 
(165) y = 42, 
be the intersection CB 
of the surface with the 
plane XY. If S re- 
present the surface 
B'LD’Q, and S and Q 
being indefinitely near 
each other, D’'QG’E’ 
is the differential of the 
surface, that is, dS = D'E’G'Q, and the differential of D’E’G’Q is 
QG'SP, that is, @S = QG’'SP. But as QG’SP is indefinitely 
small, it may be regarded as a plane area, and it is obvious that 

dxdy = HK, is its projection on plane XY, and similarly, 
dxdz = its projection on plane ZX, and 
dydz = its projection on plane ZY. 

But, as is shown in books on Analytical Geometry, the square 
of an area in space equals the sum of the squares of its projections 
on three co-ordinate planes. Hence, 

(166) (PQG'S)? = (@S)? = da*dy® + da*dz® + dy*dz?, 

Divide this by dx*dy’*, and extracting the root, we have, 

(167) i Ee nde: (: pl weak ais ): 
dxdy dy? re 

This is the second differential of the surface. 

The partial differential coefficients under the root can be obtained 
from (164) in terms of zand y, Hence the second side of (167) 


1 


242 INTEGRAL. CALCULUS, 


is a function of x and y. Represent it by x(a,y), and we may 
write (167) in the form, 

aS 

(169) eae n(X,y)-» 

It is obvious, that in the surface, « may vary, while y remains 
constant, and conversely. Hence we may integrate (169) for one 
of thesco-ordinates x,y constant, and the other variable. Integrate 
it for y variable, and we have, 


(170) Ba S (ey) dy =a'(ey) 

This may be corrected by being taken between any given limits, 
as y= 0, and y = qe. [See (165).] By these limits, (170) 
becomes, 


(170a) 2 = n'(Xy)2). 
This may be integrated for the variable x, and we have, 
(171) Ss = f[ x'(2,92) da = n''x. 


This may be corrected by being taken between any given limits, 
as «= 0,x = AB. In(171), which gives the surface CLB, we 
obviously have the surface as a function of x. By integrating (169) 
first for x variable, we would get the surface as a function of y. 

Ex. 1. Find the surface of the wedge of Example 1, last 
Proposition, 

Referring to fig. 88, we have for the equation of the surface, 


(a) 2 + y* —pa, 
and for the equation of thé plane AHB, 
(b) YY = AX. 

From (a) we get 

2 2 2 
(c) ay — Ut hye gis, and ax! — ee ete A 
dx? 4px — 4y? dy? px — y¥? 

These values put into (167), we get for the particular form (169), 

(d) nA a OR 


dxdy 2 (pax — y*)t 
Integrate this for y variable, we have, 


LENGTH OF CURVE OF DOUBLE CURVATURE. 243 


dS (4px + p?)! 1 4nx+p*)* . 0 
(ey S_ (pr + P’) f ae = CRT PY aint _ 9 
dx ~ (px —y")* “ Vv px 
This corrected by the limits y = 0, y = az, we have, 
S Ape + py? Po 
(f) ds — (4p2 FP) sin.—' a \/ d 
dx 2 Fe 


P 
Multiply by dz, and integrate, and we have, 


() 8H fi CT PY ae sina \ /=, 
; p 


a function of x, which may be integrated by parts. Hquation (g) 
corrected by the limits x = 0, 7 = Ey gives the whole convex sur- 
a* 

face of one-fourth of the wedge. 

Ex. 2, Determine the surface of the cylindric wedge of Exam- 
ple 2, last Proposition. 

Ans. S = 22RM — 4RM, 

where M = axis of the cylinder, and R = radius of its base. 
Since 2xRM is the surface of the cylinder, 4RM must be the con- 
vex surface of the parts cut off in forming the wedge, which is 
therefore quadrable. 

The last two Propositions are particularly applicable in finding the 
volume and surface of twisted surfaces. 


PROPOSITION XI. 


Determine the length of a curve of double curvature. 

Let (x,y,z) be any point P on a curve of double curvature. Let 
S be the length of the curve measured from any given point. Then 
if the curve be regarded as a polygon of an indefinite number of 
sides, we may represent one of these sides by dS, and taking dS as 
the diagonal of a parallelopipedon, whose edges are dx, dy and dz, 
we have, by the relation between the diagonal and edges of a 
parallelopipedon, 
(172) dS = (da? + dy? + dz®)*. 

Integrating this, we have, 


na 244 INTEGRAL CALCULUS. 


" (179) S ey ina + dy? + ds*)*. 
He If now the equation of the curve be represented by 
(174) v= %, y= 2%, 


it is obvious that the values of da and dy deduced from (174), will 
i be functions of z. Hence by means of (174), (178) may be ex- 
pressed in terms of a single variable, and integrated. 
Ex. 1. Determine the length of the curve, 
fl (m) 2 = 42%, yi bx, 
Fil Differentiating equation (m), we have, 
sual (n) le eAteeds, and dy? — 4b*z*d2?. 
| These values put into (173), it becomes, 
(P) S fil ue 02"\*dz, 
where c? — 4a? + 457. 
Equation (p) is integrated at (79), where a = 1, and B? = c’. 
Ex, 2. Determine the length of the curve, 
PLEO ay zig? eRe. 


PROPOSITION XII. 


Determine the integral by means of elliptic or hyperbolic 
arcs, 
Suppose we have to integrate the expression, 
(175) RE iia 
(Pa?—a?—Q)# 
If we take as the equation of the ellipse, 
. (176) avy® +. Dial a", 
\ a the differential of the arc of the ellipse is, by (119), 


(177) do lO mel Bebe 


a (a — 2°) 
If we assume 
(178) [at — (a? — B*) a*}*§ = aa’, 
hia and solve this for x, we have, 
| a (a? mom '2)8 


179 41) bee ——— : 
( ) (a* — b*)* 


ELLIPTIC OR HYPERBOLIC ARCS. 240 


Put the values of 2 and dx from (179) into (177), and we have, 
ta Rod 
[ ( a? + b?) 2? — xt — a2b?}} 

This is the same in form as (175). Hence (175) is the are of an 
ellipse. To find the axes of the ellipse, equate the like powers of x 
under the radicals of (180) and (175), and we have, 

a tf fF, and oP =a Q; 
from which the axes a and b may be found. The abscissa of the 
extremity of the elliptic arc (175), is the second side of (179). 
Again, suppose we have to integrate 
Adz 
x* (Px? — x* — Q)? 


Assume the numerator of (177) 


(182) fat — (ae) | 


(150) dz = 


SS 2 a 


x Sas = 
a —~ = 
—— SS ae a 


a 


(181) 


From which we get, 


i=) 
nw 
“™ 
oe 
to 
o 
i) 
“~~” 
Ri- 


z' (a? es b?)* 

This value of x, put into (181), changes (181) to 
(184) Ngee orirasuee: pet, a°b'dz' bossa 
x’? [ (a? + b*) 2? — x'*— ab? ]* 

Whence we see that (181) is anelliptic arc, the abscissa of 
whose extremity is the value of x in (183), and whose axes are 
found as before. 

Again, suppose we have to integrate, 

(185) Aaw’dz * 
(Px? + 2*—Q)* 
If we take the equation of the hyperbola, 
(186) vy — Px = — ab’, 
the differential of the arc of the hyperbola is, by (119), 
[ (a? + 6’) x — a‘|'dx 


(187) dz = 
a (x? — a’)* 
Assume 
(188) (a? + B*) 2? —at = aa’, 


23 


246 INTEGRAL CALCULUS. 


iy: and (187) becomes, in terms of 2’, 

| rae 9 ol 

[@oes pea 

which is of the same form as (185). Hence (185) is the arc of a 


ihe hyperbola, whose axes are found as in the elliptic arc (180). 
Again, suppose we have to integrate the form, 


Adz 


(189) BEE ati bed WE a 
| a (Pa? owl et Q): 
i im . Assume 
ine ; : 22 
ae (190) t+ Watt FT 


and (187) becomes of the same form as (189), which is therefore 
the arc of a hyperbola determined as before. 


PROPOSITION XIII. 


Determine the integral of a differential equation of the 
first order and degree, containing two variables. 

We have hitherto integrated particular expressions, or those ex- 
plicit equations in which only one side required attention. We now 
propose to integrate implicit differential equations of two variables. 

An implicit differential equation is one involving the variables and 

_ their differentials in any manner. 

The arrangement of the terms and factors of a differential equa- 
tion in such a manner that one side of the equation contains only one 
of the variables, and the other the other, is called the Separation 
of the Variables. 

To integrate a differential equation, separate the variables, and 
integrate each side of the equation. ‘The following are some of the 
cases in which the variables can be separated. 


Case 1. 
Let the equation be of the form, 
(191) Fy.dx + fx.dy = o. 


DIFFERENTIAL EQUATIONS. 247 


This is at once separated by transposing and cividing by Fy.fx, 
by which we have, 
fe Fy 
where each side is to be integrated separately. 
As an example, take the following equation, 


(a) aydx — 2°*dy = 0. 
This at once becomes, 
(0) Seok 
; oz J 
Integrating both sides of (6), we have, 
— 2. log.y +C 
2 
Case 2. 
Let the equation be of the form, 
(193) fe.Fy.dx + f'z.F'y.dy = o. 
This is separated by dividing by Fy,f’a, by which we have, 
(194) last A 
. ‘we Fy 


As an example, take the following equation. 
cx*ydx — (x — a) y’dy = 0. 
Divide by y — (2 — a), and the variables are separated, after 
which each side is readily integrated. 


Case 3. 
When the equation is homogeneous in respect of the variables. 
they can be separated. Take the form, 
(195) xy"dx + Ax y*dy = 0, 
inwhichn+m=atc=p. 
To separate the variables of (195), assume the equation, 


(196) ics £2, 
und putting this value into (195), it becomes, 
(197) xPz™dx + Ax?a'dy = o. 


Divide this by x”, substitute for dy its value from differentia- 
ting (196), and we have, 


ae 
pee 


iN. 


bh 
g 
i 


i 
VF) 
/ 


248 INTEGRAL CALCULUS. 


; (198) z™daz + Az (xdx + rdz) = 0, 
or, by multiplying and transposing, 
(199) (2” + Az°t!) dx = — Az'adz. 


From this we have, 
Zeer g™ + Agetl ” 
each side of which may be integrated separately. 
Instead of (196) we might obviously put « = zy, and obtain a 
similar result. 
As an example, take the following equation : 


tH (200) dx dg aals Az‘dz 


ai (d) 2rydx + ydy — a*dy = o. 
, This is homogeneous, the degree of homogeneity being 2. If 
we put 
(e) vie, 
then dy = zdx + adz, and (d) becomes, 
dz _ dz (1 — z*) 
1) a Sas ay 
If we put x = zy, (d) becomes, 
(g) dy —_—s-_ Qadz 
oy y TL Aces 
which is a simpler form. The integral of (g) is, 
(h) log.y = — log.(1 + 2°) + log.c et log (,—..); 
Removing the logarithms, and restoring the value of z, (h) 
becomes, 
(k) x? + y*? — cy = 0, acircle. 


Ex. 2. Integrate xdy — ydx = (2? + y?)da. 
Ans. 2? = 2cy + cc. 
Ex. 3. Integrate y°dy = 3rydx — a°dy. 
Ans. -4f == 227 Cys. 
Case 4. 
An equation of the form, 
(201) (a + bx + cy) dx = (m + nx + py) dy, 
may be rendered homogeneous, and integrated by assuming 
ee= + hh, and y=y tk. 


DIFFERENTIAL EQUATIONS. 249 


Putting these values into (201), it becomes, 
(202) (a+bh+ck+ba'+cy')d2' = (m-+-nh +pk+nzx' + py’) dy’. 
As we introduced two indeterminate quantities, 4 and k, we may 
assume in (202) any two arbitrary conditions to determine these. 
Assume the conditions 
(203) a+bh+ck=o0, and m+ nh + pk =0., 
Solve equations (203) for & and k, and substitute these values into 
(202), and it becomes 
(204) (bx' + cy') dx' = (nz’ + py’) dy’, 
a homogeneous equation, which may be integrated by Case 3. 


Case 5. 

Let the equation be of the form, 
(205) dy + Pydz = Qdz, 
in which P and Q are functions of z Assume y = 2R, R being a 
function of zx. Differentiating this, we have, dy = zdR + Rdz. 
These values of y and dy put into (205), it becomes, 
(206) zdR + Rdz + PRzdx = Qdxz. 

The quantity R being an indeterminate coefficient, we may deter- 
mine it by assuming any arbitrary condition involving R. To re- 
duce (206), assume 


(207) Rdz + PRzdz:= oa. 
This reduces (206) to 
(208) zdR = Qdz. 


Equation (207) may be divided by R, and the variables being 
separated, we have, 


(209) da =—Pdzr, -- logz= a Pdz, 


or, passing to numbers, [see equations (15), (16),] 
(210) za ews Paz, 
This value of z put into (208), we have, 
(211) OR == 62 FU aay o's =f e?*Qdx + C. 
The values of R and z from (211) and (210), put into the equa- 
tion y = zR, we have, 
(212) i= oe ( fe’ Qdx + C), 


which is the integral of (205). 
23 * 


50 INTEGRAL CALCULUS. 


The form (205) is in some books called a Linear Equation, be- 
cause it contains y only in the first power, 
Ex. 1. Integrate the equation, 


Bie (a) dy + ydx = ax*dz. 

He Here P = 1, Q = az’, and (212) becomes, 

bed (d) y = a (x — 327 + 6r — 6) + Ce, 
4 the integral of (a). The form, 
i (213) y”—'dy + Py"dz = Qy'dz, 


it may be reduced to the form (205) ; for dividing (213), by y*, and 
| Re putting y"-" — (m — n) z, (218) becomes, 


t (214) dz + (m—n) Pzdz = Qdz. 
Ri or putting P’ for (m — n) P, (214) becomes, 
(215) dz + P’zdx = Qdz. 


This is the same form as (205), 
Ex. Integrate the equation, 
(c) dy + ydx = xy*dx. 
This is of the form (218), and the integral is, 
UR) aes 1 
Fz eee, ete =e 
Suppose we have to integrate the form, 


(216) x"dy + by?'dx = gx"dz. 
This equation may sometimes be rendered homogeneous by 
assuming 
(217) = 2%, + dy = cz'dz, and y? = 2°, 
ie! These values put into (216), it becomes, 
ae (218) ex"n—'dz + ba?dx = gu™dz. 
a) i, This is homogeneous if 
ie A (219) n+e—l= Cp. =) Mm. 
| If n = 0, m = 0, and p = 2, (216) becomes, 
Hn (220) dy + byfde = gdz, - de = _Y 
oa) g— by 


where the variables are separated. 
If n = o, and p — 2, (216) becomes, 

(221) dy + bydz = grdzx. 

This is called the equation of Riccati. 


GEOMETRICAL APPLICATIONS, 251 


For (221), the conditions (219) become 

(222) e—1=2c=m -°*» c=—I1, and m= — 2, 
Hence (221) may be rendered homogeneous, if m = — 2, 
3esides the particular case m = — 2, the variables in (221) may 


be separated for some other values of m, but we will not investigate 
the subject at present. Equation (221), and the more general form, 
(216), has occupied much attention among writers on the Integral 
Calculus, but no general method of separating the variables in (221) 
has yet been devised. 

Before proceeding further with differential equations, we will give 
a number of Geometrical Applications in which such equations are 
employed. And in order that the student may become familiar with 
these equations, and with the application of the Integral Calculus to 
Geometry, we will commence with very simple problems, and after- 
wards proceed to those that are more difficult. 


APPLICATION Ist. 

Determine the curve whose subnormal is constant. 

Put the value of the subnormal, [given in Differential Calculus, 
Proposition VI., (78),] equal to a constant quantity m, and we have, 
(223) weg. 5 3 Me 

dz 
This is the differential equation of the problem. 
Separating the variables, we have, 


(224) ydy = mdz, 
and integrating, we have, 
(224a) y? = 2mer + C. 


Hence the curve possessing the proposed property is the parabola, 
with the origin on the axis of abscissas. The constant C, in (224a), 
need not be determined for the solution of the Proposition. It 
merely defines the position of the origin, and the curve, whatever C 
may be, is a parabola. ‘To determine C would require another 
condition. 


APPLICATION 2d. 
Determine the curve whose subnormal varies as the 
square of the abscissa. 


252 INTEGRAL CALCULUS. 


Using the value of the subnormal given in Differential Calculus, 
equation (73), 
(225) ydy sgh) © ra 
) dx 
nh is the problem put into equation. 
i Separating the variables, and integrating, we have, 
4 s ye me 
ie (226) FHqts 
the curve required, which is the semicubical parabola. The con- 
stant C is not determinable by the given condition of the problem. 
It merely fixes the position of the origin. 


APPLICATION 3d. 


Determine the curve whose subnormal varies as a given 
function of the abscissa. 

Denoting the given function by a, we have, by the condition of 
the problem, 

297 ydy 
(227) ae 
the differential equation of the required curve. Integrating, we have, 
(228) y= 2 f oxda, 
where, for gx, we may take any combination of x, 

If ga = mx*, (227) becomes (225), and the curve is the semi- 
cubical parabola (226). 


= 07, 


APPLICATION 4th. 
Determine the curve whose subnormal varies as a given 
function of the ordinate. 
a ha Denoting the given function of the ordinate by oy, we have, 


ent ne I Se Marine ne lt toni = —_ = P . 
RE ES es OE eg es ae = 7 — ae — a x - 
—_ er Sg ia o> a yee = nine a = : - = 
oe eS a ae rer po pt a a ape as 2 ans eo ee 
+ -- — fae Rg OR SE 
a 


| ydy 
| 229 —— = $Y; 
Hie for the differential equation of the curve. 
(230) 5 fd =x +6, 
ey 


where for gy we may take any combination of y, 


GEOMETRICAL APPLICATIONS. 953 


If gy = my*, (230) becomes - log.y = « + C, which is the 


logarithmic curve. 


APPLICATION 5th. 

Determine the curve whose subnormal varies as a given 
function of the co-ordinates of the normal point. 

Denoting the given function of the co-ordinates by 9(x,y), we have, 
yay 2(x,Y)s 
dx 
for the differential equation. ‘To integrate this, take as a particular 
case, 9(2,y) = ma*y*, and (231) becomes, 


(231) 


13 
dy —mxrdxr, °°. — tb Mba as + C, is the curve. 
y y 3 
If o(x,y) = m (x + ny), (231) is, 
(232) y a = m (x + ny), 
& 


a homogeneous equation, which may be treated by Case 3, last 
Proposition. 

If in the last five applications, we take the subtangent instead of 
the subnormal, we find curves whose subtangent possesses given 


properties. 


APPLICATION 6th. 
Determine the curve whose subtangent is constant. 
Take for the subtangent the expression given at equation (69), of 


the Differential Calculus, and we have, 
(233) seal =, i. 
dy 
for the differential equation of the curve. 
Integrate (233), and we have for the curve, mlog.y = x + C, 
Hence the logarithmic curve has a constant subtangent. In like 


manner, determine other curves whose subtangent has given properties. 


APPLICATION 7th. 


Determine the curve whose normal is constant. 
S 


254 INTEGRAL CALCULUS. 


Employing the value of the normal line given at equation (75) 

of the Differential Calculus, we have, 
2\} 
en (y+ 2) =o 
dz’ 

for the differential equation of the curve. 

From this we readily get, 
(235) dx = (m> — y*)*ydy, +. x = (m?—y’)* + C, 
or, (2 —c) + y? = m’, which shows the curve to be a circle with 
the origin on the axis of abscissas. 


APPLICATION 8th. 


Determine the curve whose normal varies as a given 
function of the ordinate of the normal point. 

Denoting the given function of the ordinate by gy, we have, 
dy’ \= 
(236) (y 4+ y? ai ) = 9Y; 
for the differential equation of the curve. From this we get, by 
separating the variables, 
(237) dx = ydy[ (gy)? — yy *. 

By putting for gy any combination of y, this may be integrated, 
and the curve determined. 


APPLICATION 9th. 


Determine the curve whose tangent line is constant. 


Using the expression for the length of the tangent line, at (74), 
Differential Calculus, we readily form the differential equation. 


APPLICATION 10th. 


Determine the curve whose tangent line varies as a 
given function of the ordinate of the point of tangency. 

Denote by gy, the given function of the ordinate, and employing 
the length of the tangent at (74), Differential Calculus, we readily 
form the differential equation. 


GEOMETRICAL APPLICATIONS. 255 


APPLICATION 11th. 
Determine the curve whose normal equals the abscissa 
increased by the subnormal. 


: ee ( dy : 
238 gt fog oe maha 8 vy. 
Oye) y elaai ies az 
is the differential equation. From this we have, 
(239) y'dxz — 2xydy — x*dx = o. 


This is a homogeneous equation, which may be integrated by 
Case 3, Proposition XIII, and from which we have, 
(240) y? + 2? — 2x = o, 

This is a circle with the origin on the perimeter, and centre on 
the axis of 2. 


APPLICATION 12th. 
Determine the curve in which the subnormal is to the sub- 
tangent as the ordinate is to the abscissa of the point of 


tangency. 
By the problem, we have, 
(241) ydy : Phe Ss tse e 
dx dy 


from which we get, 
(242) y ‘dy — x “da, 
the differential equation. Integrate this, and we have, 
™M 
eye 
for the curve. Clearing this of radicals, we have a parabola with 
the co-ordinate axes tangent to the curve, and the origin at the in- 
tersection of the axis and directrix. 


(248) y= eit 


APPLICATION 18th. 


Determine the curve in which the subnormal isto the sub- 
tangent. as the abscissa to the ordinate of the point of 
tangency. 

By the problem, we have, 


Se Se 


etm 


ees 


256 INTEGRAL CALCULUS. 


d ydx 
244 Oe ee 
(244) is iy y 
from which we have, 
(245) y dy =2dz, .:. y2 =x? + C, 


fi which is a line of the 6th order. 


a APPLICATION 14th. 
| te Determine: the curve in which the area varies as the 


| FY. ordinate. 
as Take the value of the area at (113), Proposition I., and we have, 
| is (246) S ydu = tty, 


for the equation of the curve. 

In order to integrate an equation involving an integral, as (246), 
we must dispose of the sign of integration by first differentiating 
the equation. Differentiate (246), and we have, 

(247) ydx = mdy. 
This is the differential equation of the curve required. 
Integrate this, and we have, 

(248) x= milogy + C 

for the curve required. 


APPLICATION 15th. 


Determine the curve in which the area varies as a given 
function of the ordinate. 


Denote the given function of the ordinate by ¢y, and we have, 


(249) J yde = oY; 
for the equation. Differentiate this, and we have, 
(250) ydx = 9'ydy, 


im where 9’y is put for what gy becomes when differentiated. 
Separate the variables in (250), and it may be integrated for any 
given value of oy. 


APPLICATION 16th. 


Determine the curve whose area varies as the product 
of the co-ordinates. 


GEOMETRICAL APPLICATIONS. 257 


sy the problem, we have, 


(251) Qs ydx = mxy, 
for the equation of the curve. Differentiate this, and we have, 
(252) (y — my) dx = mady, 


which belongs to Case 1, Proposition XIII. 
Separate the variables, and we have, 


(258) A woh,» dy ee tlomei== gh as log.cy, 
x I—m y 1—m 
or removing the logarithms, {see equations (15), (16),] we have, 
(254) aim — (cy). 
If m = 2, this becomes, 
(255) «= (cy), 


the common parabola. 


APPLICATION 17th. 
Find the curve in which the area varies as the sum of the 


co-ordinates. 
Express this law of variation by m (a + y), and we have, by 
the problem, 


(256) Sf ydu = m(xr + 9), 
for the equation. Integrate this, and we have, 
257) x = m log.(y — m) + C, 


for the curve. 


APPLICATION 18th. 
Find the curve in which the arc varies as the 3 power of 
the abscissa. 
Take the general form, for the length of an arc, at Proposition III, 
(119), and we have, by the problem, 
t 3 
(258) SJ (dx + dy’) = mz?, 
Differentiate this to dispose of the sign of integration, and we have, 
(da? + dy?) = 3 mada. 
Squaring, transposing and extracting the root, we have, 


(259) dy = ee mee — 1) de. 
24 


208 INTEGRAL CALCULUS, 


Integrate this, and we get, 
8 9 3 
260 Y= < ma — 1) 
(260) y=55 (4 + ©, 


i the semicubical parabola. 


APPLICATION 19th. 


nih Determine the curve in which the arc varies as a given 
By function of the ordinate, or abscissa, or of the co-ordinates. 
By the Proposition, we have, 


L (261) Cas: ae ee 
Se (262) SJ (de? + dy’)! = ox, or, 
(263) J (da? + dy?) = o(x,y), 


according as the arc varies as a function of the ordinate, or ab- 
scissa, or co-ordinates. 

For any given value of oy, or 2, or 9(x,y), these equations can 
be differentiated, to dispose of the sign of integration, and if the 
variables are then separable, the integration may be effected, and 
the curve determined. 


APPLICATION 20th. 
Determine the curve whose surface of revolution varies as 
a given function of the ordinate. 
In Proposition IV., (127), we have the expression for the surface 
of revolution. Putting that expression equal to a given function of 
_ the ordinate, which we will denote by gy, we have, — 
(264) . Qn fy (dx? + dy?)* = Qy. 
Differentiate this, and we have, 
fe (265) 2ay (da* + dy?)* = g'ydy. 
A From which we get, 
) (266) dx — L(0'y)* = An’yT dy 
2ny 
This may be integrated for any particular value of oy. 


GEOMETRICAL APPLICATIONS. 259 


APPLICATION 21st. 

Determine the curve whose volume of revolution varies 
as a given function of either or both of the co-ordinates. 

By Proposition V., (130), we have the expression for the volume 
of such solids. Then by the problem, we have, 
(267) af ydx aes, x fydx Oy; af ydx = $(2,y). 

These equations may be integrated for any particular forms of 
the given functions. 


APPLICATION 22d. 


Determine the curve whose polar subtangent varies as a 
given function of the radius vector. 

In Differential Calculus, Proposition XXXIIL., (172), we have the 
general form of the polar subtangent, which put equal to gr, a given 
function of the radius vector, ae we have, 


(268) mda er, et Cae 
dr re 
which for any particular value of gr, may be integrated. 

If gr = mr’, (268), becomes, 
(269) dy = mdr, + e=mr4+C, 
which is the spiral of Archimedes. 


3 


APPLICATION 234. 


Determine the curve whose polar area varies as a given 
function of either or both of the polar co-ordinates, 

By Proposition VI., (133), we have the polar area, and by the 
ey we have the equations, 
270) af Mde = = da, +f rdw = or, af rds = (7,0). 

These equations may be integrated for any particular values of 
the given functions. 


APPLICATION 24th. 


Determine the curve which cuts the radius vector at a 
given angle. 


260 INTEGRAL CALCULUS. 


By Proposition D, Appendix to Differential Calculus, we have the 

i value of this angle in terms of the tangent, sine, and cosine of the 
he angle, Hither of these put equal to a constant, and integrated, gives 
Bre the curve. If we take the tangent, then putting m for the constant, 


Bi we have, 
By 
ait ~ rd 
f (271) "= mM, 
We dr 
: for the differential equation. Integrate this, and we have, 
} 


ers the curve required, which is the logarithmic spiral. 


n) APPLICATION 25th. 

Determine the curve which cuts the radius vector at an 
angle whose tangent, or sine, or cosine, varies as a given func- 
tion of either or both of the polar co-ordinates. 

By the same Proposition D, we have the values of these trigono- 
metrical lines, any one of which put equal to a given function of 
one or both of the polar co-ordinates, furnishes the differential equa- 
tion, whose integral is the curve required. 

If, for example, the tangent varies as the variable angle, we have, 


rdw 
(273) Se me Wy 11 FE 
1 hy dr 
for the differential equation. 
~ da dr : 
(274) et Ie ee MOS wee at IOP Oh, OF Been ee 
ro) r 


APPLICATION 26th. 
at Determine the curve which cuts a given family of curves at 
a ie a given angle. 


We have already spoken 
of a given family of curves 
as a system of curves formed 
by giving different values to 
a parameter that enters the 
equation of acurve. Thus 
in the equation of the parabola, 


GEOMETRICAL APPLICATIONS. 261 


(275) yy =a px, 
if p take different values, (275) expresses different curves, but all 
of the same family or species, viz : parabolas. 

Let the equation of a family of curves be represented by, 
(276) $(x,y,8) = 9, 
where 4 is a parameter that enters the curve. Let DH be one of 
the curves of this family. Let SE be the curve which cuts DH at 
a given angle. Draw PN and PL tangent to these curves at the point 
of intersection P, Put V for the tangent of the given angle NPL, 
and suppose, 
(277) J(x',y’) ==, 
be the equation of SH, the curve required. 

Put p and p’ for the differential coefficients of (276) and (277), 
respectively, then by Trigonometry, we have, 


(278) Rest DR SAUP. 
1 +pp 


The value of p involves the 2,y and g of (276). Hence (278) is 

a function of x,y,6,p’, and may be represented by 
279) Vis FE (2,958;p)). 

Eliminate 8 between (279) and (276), and we may represent the 
result by 
(280) V =f (2,y,p'). 

But the a and y of (280) appertain to the point of intersection P, 
and are consequently the same as 2’,y’, the co-ordinates of SH, the 
curve required. Hence (280) may be written, [putting for p’, its 
equal dy’ ~ dz’), 


(281) V= f ("9 hy 


az’ 


This is the differential equation of the curve required. Its in- 
tegral is represented by (277). The curve (277) is called The 
Trajectory of the family of curves (276). 

If the angle of intersection P be a right angle, (278) becomes, 
(282) 1 + pp’ = 0, 
and the trajectory is called orthogonal, For the orthogonal tra- 
jectory, proceed with (276) and (282) as above directed, for (276) 
and (278). D4 ¥ 


SE eri. Sepa 


al 
sf 
4 


i Oe PS 


eer 


< 


= 


-_—_—_ 


4 
: ee a = Se < L. bie 

Sy arg _ ot : ~~ rey ang eo a ; siecen ape 
tages ot: eee Se Dee eet 

ea ; Te ee eh ee ae ee re Rae ee 


262 INTEGRAL CALCULUS. 


We will add several examples of this Proposition. 
Ex. 1.—Determine the orthogonal trajectory of the curve, 
(a) = ‘pz. 


Differentiate this, and we have, a = 262 == Dp 
x 


This value of p put into (282), we have, 
(5) 1 + 26xrp' = o. 

Eliminate 6 between (a) and (b), and we have, [putting a’ and y’ 
for x and y], 
(c) x'dz' + 2y'dy' = o. 

The integral of this is, x” + 2y— C, an ellipse. 

Ex, 2.—Determine the orthogonal trajectory of the curve, 


(d) y = Bx". 
The differential of this is, p= nga*-', This put into (282), 
we have, 
(e) 1 + 162 "'n' = 0, 
Eliminate p between (d) and (e), 
(f) “. gw'dz' + ny'dy' = 0, ony 


an ellipse. If nm = 1, (d) is a family of straight lines, and (f) is 
a circle. 


Ex, 3.—Determine the trajectory which cuts, ata given angle, 
the family of curves. 


(2) y = av, 
Differentiate this, and we have, p= a— ”. This value of Pp 
Hb 
put into (278), we have, 
(h) V = E&Y. (Ve 6 de! = (o! — Vyay 
ope ee AE 


This is a homogeneous equation, which may be integrated by 
Proposition XIII., Case 8. This example is evidently the same as 
Application 24th, above, 

Ex, 4.—Determine the orthogonal trajectory of the family of 
curves, 

(k) y? = 2rz — x’, 
which is a family of circles tangent to the axis of y at the origin, 


DIFFERENTIAL EQUATIONS OF THE HIGHER DEGREES. 263 


From (k) we have, for the differential coefficient, 
(l) ge r ; x 
which put into (282), we have, 
(7) y+(r—x)p =o. 

Eliminate 7 between (k) and (m), and we have, [after putting 2’ 
and y' for x and y], 
(22) 22'y'dz' + (y? — z')\dy'. =. 0; 
which is a homogeneous equation, and integrated at Proposition XIIL., 
Case 3, from which we have, for the required trajectory, a circle 
passing through the origin, and tangent to the axis of 2. 

We will give other Geometrical Applications under other Proposi- 
tions, preserving the order of the number of the applications. 


PROPOSITION XIV. 


Determine the integral of differential equations of the 
first order, exceeding the first degree. 
The most general form of a differential equation of two variables, 
and of any degree 2, is 
(283) dg sings cama otek ol! hha Wehtoe 
dx” 7 agi dx 
where M, N, and P are functions of # and y. Solve (283) for 


? 


= and representing its roots by p, p’, p’’, &c., we may divide (283) 
da 

into the several simple equations, 

GH a iy a= 4 "Set Og pee 
dx dx 

The integral of any one of (284) is the primitive of (288), which 
consequently has 7 primitives. 

Though many of the Applications given under last Proposition 
involved differential equations of the second degree, yet as the inte- 
grations under those Applications were very simple, we did not deem 
it important to distract the attention of the student by introducing 


into those Applications the consideration of the differential equations 


‘p= 0, We. 


ey 


a oe me 


264 INTEGRAL CALCULUS. 


exceeding the first degree, especially as the results in the Applica- 

tions given would be, in general, the same. 

the present Proposition by an example or two. 
Ex, 1.—Integrate the equation, 

(a) dy and". 
Here the two roots of the differential coefficient are + m, and 

— m, hence (a) is resolvable into the two equations, 


We will now explain 


(b) dy = mdx, and 
(d) dy = — mdz, 
The integrals of these are, 
(e) y=me+C, and 
(f) y=—mz + C. 
Either of the equations (e) and (f ), is a primitive of (a). 


FIG. 91, 


Equation (e) designates a line BC, whose 
ordinate at the origin is AB = C, and (f) 
designates a line EF, whose ordinate at the 
origin is AE = C’, If C=C, (e)and (f) 
designate the two lines BC and BD, having a 
common ordinate AB at the origin, and 
making equal angles with the axis of y. 
Hence we may add the same constant C after the integral of each 
of the equations (284). 

If we multiply (e) and (f) together, we have, making C = C’, 
(g) (y — mx —C) (y + mz — C) = 0, 
which single equation represents the complete integral of (a). 

The same thing may be represented by putting (a) in the form, 
(i) dy =.= nda, epg [ete BE 
which represents two lines BC and BD, equally inclined to the axis 
of y, at the same point B. In the same manner, if each of the 
equations (284) be integrated, and the same constant added to each, 
the product of these integrals is the complete primitive of (283). 

Ex. 2.—Integrate the equation, 

(k) ayes wad x: 

The two integrals of this appertain to the two branches of the 

semicubical parabola. 


DIFFERENTIAL EQUATIONS OF THE HIGHER DEGREES. 269 


If we could always solve (283) for the dif fferential coefficient, the 
foregoing process would suffice to integrate equations exceeding the 
first order. But (288) cannot in gener ral be solved, when it exceeds 
the fourth degree, and even in those cases where it is resolvable, the 

radicals are often so complicated that the integral of the partial 


ess co: 84) cannot be obtained. The solution and integration 
of (283) can, however, be effe scted in the following cases. 
Case 1. 


If the equation (283) does not contain y, and can be readily solved 


for a, it can be integrated as follows. 
Put p for the differential coefficient dy ~ da, and if (283) be 


solved for a, we may represent the result by 


(285) mi OH 

Integrate by parts, the equation dy = pda, and we have, 
(286) y= PL — f «dp. 

By means of (285), this becomes, 
(287) yo 7D ett AS dp. 


After integrating the last term, of (287), the coefhicient p may be 
eliminated between (285) and (287), and we have the integral 
required. 

Ex.—Integ rate 2 +mp=n(l+p oF 

This is readily solved for 2, and may be integrated as directed in 
(285) to (287). 

If the equation (283) canrfot be solved readily for either x or p, 
put p = ua, as in the example, 


(l) zt p® = mpr 
Solve this for 2, and we have, 
mu mu? 
Sas andi jp) toss 
1+ wv 1+u 
Put these values of x and p= into the form, 
(m) y = f pdx, 


which is the integral of the assumed equation, dy = pdx, and we 
have, 


(n) y= nt 4 i *du (1 — 2u’) oy 
(_1 + u?) 


266 INTEGRAL CALCULUS, 


Integrate (n), and uw may be eliminated between it and the value 
of 2 in (1), 


Case 2. 
If (283) does not contain x, and can be readily solved for y, 
represent it by 
(288) ¥y = Fp. 
Differentiate this, and we have, dy = dFp = pdz. Solve this 
for dx, and we have, 


(289) peng D ni stsee ky say faa 
p Pp 


Eliminate p between this and y = Fp, and we have a relation 
between y and «x, which is the primitive of (288). 


Case 3. 

If the equation be homogeneous in respect of x and y, the varia- 
bles may be separated as in the case of homogeneous equations at 
Proposition XIII., Case 3. For if we assume, 

(290) Vise ea, 
and substitute into the homogeneous equation, x will disappear from 
that equation. If the equation can then be solved for z, we have, 


(291) Ge TOs aoe OTP, 
and from (290), we have, 
(292) dy = zdx + «dy = pdz. 


Eliminate z and dz from this last equation, by means of (291), 
and we have, 
(298) OP witty 
Nees oat 
Eliminate p between the integral of this and the equation, 
y = xfp, [from (290,) (291),] and we have a relation between x 
and y, which is the integral required. 


Ex,—Integrate the equation, py = nx (1 + p*)’. 


Case 4. 
If the equation be of the form, 
(294) y = px + Fp, 


DIFFERENTIAL EQUATIONS OF THE HIGHER DEGREES. 267 


it is always integrable. Here Fp denotes a function of p, but con- 
tains neither x nor y. 
Differentiate (294), and we have, 


(295) dy = pdx + adp + F'pdp. 
But since 
(296) dy = pdx, 
subtracting, we have, 
(297) o == (x + F'p)dp. 
This gives the two conditions, 
(298) dp = 0, and x2 + Fp = 0. 
The first of these gives p = c. Hence the integral of (294) is, 
(299) => cx... Ve. 


Hence the complete primitive of the form (294) is obtained by 
making pconstant. Again, if we eliminate p between (294) and 
the second of (298), the resulting equation, which may be repre- 
sented by 
(300) F (x,y) aor Oh 
will be an integral of (294), but will not contain any arbitrary con- 
stant. This result is called a singular solution of (294). We will 
revert to these solutions hereafter. The form (294) is known as 


Clairaut’s Form.. The second of (298) might also be integrated 
by putting for p, its value, dy’ 
dx 


Case 5. 
If the equation be of the form, 
(301) =r 1 k's 
where P and P’ are functions of p, we first differentiate (301), and 
we have, 


(302) dy = Pdx + adP + dP’. 
Since we have, 

(303) dy = pdz, 

by subtraction we have, 

(304) o = (P —p) dx + xdP + dP’. 


From this we get, 


268 INTEGRAL CALCULUS. 


edP..<'. atid P’ 
Rp | Pay. 
which is the same form as (205). 
We will give a few Geometrical Applications involving the pre- 
ceding cases. 


(305) dx + 


APPLICATION 27th. 


Determine the curve whose tangent makes with the axis 
of ordinates an angle whose cosine varies as a given function 
of the co-ordinates of the point of tangency. 


Since the differential of the arc is (da? + dy’)*, the cosine of the 


proposed angle is, ay -, and by the problem, we have, 
(da* + dy) 
(306) nen OB al 2 (x,y). 


(da? + dy? 
If o(2,y) = ent (806) becomes, 
y 


(307) ydx = nx (da* + dy’)*, 
which being homogeneous, may be treated by Case 3, Prop. XIV. 


APPLICATION 28th. 

Determine the curve in which the rectangle of the ab- 
scissa, and abscissa diminished by a given line, equals the 
rectangle of the arc and that line. 

By the problem, we have, putting n for the given line, 

(308) x(x —n) = nf (dx® + dy’). 

Differentiate this, to dispose of the sign of integration, and 

we have, 


(309) 2adx — ndx = n (da? + dy’), 
or dividing by dx, we have, 
(310) 22 —n=n(1 + p*)', 


which may be treated by Case 1, Proposition XIV., and the curve 
determined. 


GEOMETRICAL APPLICATIONS. 269 


APPLICATION 29th. 
Determine the curve whose tangent cuts off, with the co- 
ordinate axes, a triangle of given area. 


FIG. 92. 


Let the equation of the tangent Y 
line CB be, 
(311) y—y =p(t#— 7’). Cc 
The distances from the origin to 
where this line cuts the axes of a 
and y, are found by putting first, 4 
x = o, andthen y = 9, in (811), 
from which we have, 


(AY -y'aap2! = AC) hand. — & Pee agp: 


Putting a’ for the area of the triangle CAB, we have, 
(312) (y' — p2')y = — 2pa’, 

The negative sign may be changed to positive, as it only affects 
the position of the area. Hence (312) becomes, omitting the 
accents, extracting the root, and transposing, 


(313) y= px +a J/2p. 
This is of the form (294). Hence its integral is, 
(314) y == 'cr + a) 7 2c, 


which is the equation of a straight line If (313) be differentiated, 
the particular form of (298) is, 
(315) aoa ed ay 
J/ 2p 
Eliminating p between (313) and (315), we have, for the singu- 
lar solution, 
(316) ry = — = 


2 


= 0. 


? 


a hyperbola, as in Proposition XXXVII., Example A, Differential 
Calculus, This would be the curve PS. 
If the second side of (312) be treated with the negative sign, 
we have, instead of (3138) and (315), 
T 25 


270 INTEGRAL CALCULUS. 


(317) y = pe +aJ/—2p, and x + f =o. 
Vv —2p 
If p be eliminated between these two, we have, 
2 
(318) ty : as before. 


APPLICATION 30th. 


Determine the curve whose tangent between the axes is 
always of the same length. 

Put m for the given length CB, figure 92, and using the lengths 
(A), Application 29th, we have, 


aaa \8 
i, eee ee es 
| BF 
From this we get, 
(320) - y=pe+—h" _, 
(pi + 1) 


which is of Clairaut’s form (294). Hence put C for p in (320), and 
we have its complete primitive. Differentiate (320), and the parti- 


cular form of the second of (298) is, 


m 


——; 
(p’ + 1)? 

If p be eliminated between (321) and (320), we have for the sin- 
gular solution, 


(322) y3 4-3 oe ms. 
If we put in (321), for p, its value Bs and then integrate (321), 
a 


(321) Ove + 


we find, 
(328) y = — (m3 — a3)? + C, 

To determine C, observe that when y = 0, we have x = m. 
Hence C = o and (828) becomes (322). 


APPLICATION 3lst. 
Determine the curve such that a perpendicular drawn from 
a given point to its tangent, is of a given length. 


SINGULAR SOLUTIONS. 271 


Suppose the given point be the origin, and m the given length AH, 
(fig. 92], of the perpendicular. For the tangent CB we have the 
equation, 

(324) y—y =p(r#—7’). 

For the equation of the perpendicular AH, on (824), from the 

origin, we have, 


(325) py=—e. 
For the length of the perpendicular AH, we have, 
(326) m? = x + y?, 
Eliminate a and y from these three equations, and we have, 
(327) y' = px' + m(p? + 1), 
This being of Clairaut’s form, (294), its integral is, 
(328) y = cx’ + m(c* + 1). 
The singular solution of (327) is, 
(329) xe? +. y'2 — m?, 


which is the equation of a circle. 


PROPOSITION XV. 


Determine the singular solution of a differential equation. 


We have given singular solutions of the last three Applications, 
and from the mode of determining them, it is obvious that the com- 
plete primitive being the equation of a line, [straight or curved], the 
singular solution is the locus of the intersection of this line with its 
consecutive line. For equation (315), for example, is obviously the 
same as would be obtained by differentiating (313) for « and y con- 
stant, and p variable, and the result of eliminating p between (313) 
and (315) is obviously the same as the result of eliminating c be- 
tween (314) and the differential of (314), for c variable. Hence 
the determination of singular solutions is resolved into the determi- 
nation of the locus of intersection of a line with its consecutive line, 
a subject fully investigated in the Differential Calculus, Propo- 
sition XXXVI. 

To obtain the singular solution of a differential equation, there- 


272 INTEGRAL CALCULUS, 


fore, we first obtain its complete primitive, and eliminate the arbi- 
trary constant from this complete primitive, by the principle of con- 
secutive lines. If the arbitrary constant disappears in differentiating 
the primitive for this constant variable, the equation does not admit 
of a singular solution. 

By eliminating c between (328) and its differential for ¢ variable, 
we get the singular solution (829). Consequently the curve given 
by the singular solution touches all the lines contained in the com- 
plete primitive. 

The singular solution of Application 31st is the same as would be 
obtained by the principle of consecutive lines, if the Application had 
been in these words: A line 1s drawn ata given distance from a 
given point ; determine the locus of intersection of this line with 
its consecutive line. Equation (328) is the equation of a straight 
line in terms of the perpendicular on it from the origin, and of the 
angle this perpendicular makes with the axis of x. 

Exx.—Determine the singular solution of the equation, 


(a) ady — ydx = (x* + y*)*dax. 

The complete integral of this is, [Proposition XIII., Case 8, Ex.2], 
(0) rs SCH +, 

Hence the singular solution is, 
(c) rie a point. 


For the mode of obtaining the singular solution without first inte- 
grating the differential equation, see Proposition (E), post. 


PROPOSITION XVi. 


Determine the integral of ‘a differential equation of the 
second order, containing two variables. 

The most general form in which such an equation can occur, is, 

dy dy\ _ 

(330) ew, Beet he 0, 
which contains the first and second differential coefficients, and both 
the variables. 

We will not attempt the integration of (330) in its most general 
form, but will examine several cases of it. 


DIFFERENTIAL EQUATIONS OF THE SECOND ORDER. 273 


Case 1. 
If (330) contain neither y nor the first differential coefficient, it 


becomes, 


(331) o(*, vy St) = 9, 


This solved for the second ‘ee coefficient, gives that co- 
efficient as a function of a, which denote by x2, and we have, 
dy 


Sar, 
dx? 


(332) 


This may be integrated as at (106). 


Case 2. 

If (330) contains neither x nor the first differential coefficient, it 

becomes, 
dy 

333 (: 3) = 0, 
( ) PRY: dz? 

This solved for the second differential coefficient, we have, 

dy 

334 EE A 
(334) aa ie 

Assume —* i nl = p, then by differentiating, we have, 

ax 
(335) a a tS ei 
dx* dx dy 


Equating (334) and (335), we have, 
(336) pdp=-rydy, °- p= 2fnydy=y + ©, 
where y’ is a function of y, and is put for the integral of 2 f wryly. 
Restore the value of p in (386), and we have, 


(337) ay =e tanh oy ere Ly Mr A 
dx (y — C): 


the integral required. 


Case 3. 
If (330) contain neither x nor y, it becomes, 


(338) Ame: fy )=0 
dx bet iy 


arti. | 


274 INTEGRAL CALCULUS. 
Assume ee hee ay = we These values put into 
dx dx? dx 
(338), we have. 
di 
(339) (>. *) eet (8 
dx 
an equation containing p and 2, of the first order. Solve it for dz, 
and we have, 


(340) dx = fpdp, 
(341) ¢=-f'p 40. 
But since dy = pdx, we have, 
(342) dy = p/pdp, 
(343) y=f'p +l, 
Eliminate p between (341) and (343), and we have, 
(344) F (2,y,0,C’) = 0, 
the complete primitive of (388). 
Case 4. 
If the equation (330) does not contain y, it becomes, 
(345) o( =. dy. hay a 
dx. da 
dy oe gp These values put into 


Assume ima Copy eg 
(345), it becomes, 
(346) o(=», 2) = 

dx 

which is an equation of the first order in p and 2, and may be inte- 
grated by Proposition XIV. 

Suppose its integral determined, and of the form, F (#,p,C,) =0- 

In this put for p, its value, and we have again a differential equa- 
tion of the first order, which may be treated by Proposition XIV. 
The result will be, 
(348) S (e@y,0,0') = 9; 
the complete primitive. 


Case 5. 
If (330) does not contain x it becomes, 


rs) 
~~) 
qi 


DIFFERENTIAL EQUATIONS OF THE SECOND ORDER, 


2 
349 ee ONS o 
PRY ; 
dn ads 
Since cy — i 1 and ay = pap: (349) becomes, 
dx da? dy 
ae | 
50 (v0, 2) = 0 


a differential equation of the first order in y and p, which may be 
treated by Proposition XIV. 


Case 6. 
If (830) be homogeneous in respect of the variables and differen- 
tial coefficients it may be reduced to an equation of the first order. 
3y the equation being homogeneous in respect of the differential 
coefficients, is understood that the degree of homogeneity of 


dy i. zero, and of Gu is — 1. Thus the equation, 
dx ax 


é Re dy 
351 ei +r +y—=0 
CoH) dx* J dx 4 
is homogeneous, the degree of homogeneity being unity. Assume, 
; % 
(352) dy =p, y= ux, and ay = gene 
dx dx? x 


‘These put into (330) will render (330), when homogeneous in the 
n'* degree, divisible by x”, and dividing by 2”, (330) will contain 
only the three quantities, p,w,2, and may be written, 


(353) Q(p,u,z) = 0. 
Differentiate y = ux, and we have, dy = udx + xdu; from 
this, and dy = pda, we get 
dx du 
(354) a, ey Se 
x P= adt 
But from the first of (852) we have, 
dy ~~ dp 
da® dx 
which equated with the last of (352), gives, 
l 2 
sy = —) or, dx nam dp 
dz a x % 


This equated with (854), we have, 


276 INTEGRAL CALCULUS, 


(355) OY 3 Alaa 
z p—u 

Eliminate z between (353) and (355), and the result contains two 
variables, and may be represented by 
(356) a(p,u,dp, du) = o. 

This being of the first order, may be integrated by Proposition 
XIV. Then by means of this integral, p may be eliminated from 
(354), and that equation integrated, we have, 

(357) log.2 = fu, . 

Eliminate w between this and y — wa, and we have the relation 

between z and y, 


We will adda few Geometrical Applications involving these cases. 


APPLICATION 32d. 
Determine the curve whose radius of curvature is constant. 


By equation (199), Differential Calculus, we have the length of 
the radius of curvature of any curve, which if we put equal toa 
constant m, we have, 


ates dy\2 | da 
(358) oe (+4) ete een 97h, 


This containing neither x nor y, but the first and second differen- 
tial coefficients, is a particular example under Case 8. Hence taking 
(358) with the negative sign, it becomes, in terms of p and z, 


(359) da ay ek aes 
(1 + p*)? 
Integrate this, and we have, 
ea P _40C, — Since, 
(1 +p’) 
m 
(360) y =P 4p 


Eliminate p between (359) and (360), and we have for the 
primitive, 
(361) (C—zx?+4+(C—yr=m, 
a circle, the radius of curvature being the radius of the circle, 


Ns) 
~) 
aj 


GEOMETRICAL APPLICATIONS. 


APPLICATION 38d. 


Determine the curve whose radius of curvature varies as 
the square of the abscissa. 
By the problem, we have, 
(362) —(1 ey Reh yuan vente 
dx’ dx? m 
for the differential equation, which is a particular example of Case 4. 
Substituting, as in Case 4, we have, 


en 
(363) —(1+ p’) 2dp = ma~dz, 
which integrated, [by (78),] gives, 
(364) SEO 6 Bt aC! 


(l + Lp?) x 
This may be again iterated by Proposition X1V., Case 1, and 
the curve determined. 


APPLICATION 34th. 


Determine the curve whose radius of curvature equals 
the normal. 

Putting the length of the normal equal to the radius of curva- 
ture, we have, 
= ay _ dy 
899) 4 (+5 ie — =-(1 gee dat 

This is a particular example of Case 5. tee this to p and 
y, we have, 


(366 ay =— pap” 
y ped 
The integral of this is, 
: C 
ete are Ty 
le 
Restoring the value of p, and integrating again, we get, 
(367) (ec —Cy + y? = C, acircle. 


If we take the second side of (365) positive, we get, instead 
of (366), 


INTEGRAL CALCULUS. 


(368) *=C(p* +1), 

and restoring the value of p, we get for the next integral, a logar- 
ithmic equation. Now we know, Differential Calculus, that the 
radius of curvature with a negative sign, gives a curve concave 
towards the axis of abscissas, and with a positive sign, a curve con- 
vex towards the same axis. Hence the logarithmic curve given by 
integrating (368) is convex towards the axis of x. This curve has 
many properties analogous to those of the circle. 


APPLICATION 35th. 
Determine the curve whose radius of curvature is a given 
multiple of the normal. 
Let n be the given multiple. Then putting n times the first side 
of (365) equal the second side, and proceeding as in Application 
34, we get for the final integral, 


: Soe 
(369) = =f ydy (c—y"). ? 
This fulfils the condition of integrability at (65), if n be an odd 
number, and the condition of integrability at (66a), if n be an even 
number. Hence the curve is always determinable for integer 
values of n. 


APPLICATION 36th. 

Determine the curve whose radius of curvature is a given 
multiple of the distance from a given point to the point of 
osculation. 

Let the origin be the given point, and z the given multiple, then 
by the problem, we have, 

Q- 2 A oe dy ‘Bay 
(370) n(x + y’) =—(1+ _) Pee oe 

This is a particular example of Case 6, last Proposition, and pro- 
ceeding with it as directed in Case 6, we get for the particular 
form of (356), the equation, 

(371) (d+ p’)idu =n(p—u) (1 + wu) dp, 
an equation integrated by Euler, by means of circular arcs. [Vide 
La Croix, vol. [I., 307]. 


4 


GEOMETRICAL APPLICATIONS. 279 


APPLICATION 37th. 
A body D moves with a given velocity along a given 
straight line, a body C moves with a given velocity directly 


~~ 


towards D ; determine the locus of C. 


FIG. 93 

Let D move on the axis of y, and = 
since the velocities are given, their ratio 
n is given. B 

When D is at A, let C be at B, on 
the axis of 2, and let the points C and 
D be their position at any other time. 

Put AH = y, HOC=27, BO= xz, Cc 
Then the subtangent is, 

HD = — ad , H D * 


this increased by AH, or 2, is the distance gone by D while C goes 
the distance BC = z. 
The ratio n of these distances being given, we have, 


(372) (y—=2) = ne, 
dx 
or putting for z its value in (122), 
(373) (y¥— pr) = nf (1 + p*)*dz. 
This differentiated to dispose of the sign of integration, we have, 
(374) ay — dp 


oat ies pee eS Sele 
Ho (1 as py) 
which comes under Case 4, By (63) the integral of this is, 


(375 “» (Cx)? = (1 + p*)t —p. 
To determine C, we observe that when x = AB = a, we have, 
p =o. These values put into (375), we have, C = Boe 
a 


This value of C put into (375), that equation becomes, after 
restoring the value of p, and integrating, 
; n—+-1 n 
(376) Qy) ee ae ee eC 
a” (n+ 1) (n + 1) a" 


ee > ere 


ee aaa 
th - = 


wae abe ~ 


er 
ae 


role = 
Pee ae <= 
TS Nl 


— 


280 INTEGRAL CALCULUS. 


From this C may be eliminated by the consideration that when 
y = 0, x = a, and the curve is completely determined. 
This is The Curve of Pursuit of the French Mathematicians. 


PROPOSITION XVII. 


Determine the integral of differential equations of two 
variables exceeding the second order. 
We shall attempt this in only a few cases. 


Case 1, 
If the equation be of the form, - 
n n—1 
(377) ray (ket ad cat Ds se 
dz” da 
Assume, 
(378) ay = ety any = du 
ae dx” dz 
These values put into (377), we have, 
du 
(379 F (e u ) a 
Ca dz 


which is of the first order in respect of wanda. This being inte. 
grated, we may represent the result by w = fa, and this value of 
w put into the first of (878), we get a form which may be integrated, 
and gives a relation between y and z. 


Case 2. 
If the equation be of the form, 
2s any. dy 
380 ia A 
Cad, dx” dx 
Assume, 
dy d”y au 
381 at | oe eee 
(351) Gaze ; dx” dz? 


These values put into (380), we have, 


(382) F ( u) 7 


eh 
dx? 


DIFFERENTIAL EQUATIONS OF THE HIGHER ORDERS. 281 


which is of the second order, and may be integrated by Case 2, 
Proposition XVI. If the integral of (382) be wu = fz, this value of 
u put into the first of (881), we may integrate (381), and find the 
relation between y and z, 


Case 3. 
Suppose the equation be of the form, 
(383) d*y + Pd™"'ydx + Qd"*ydz* +..-. Sydx” = 0, 
where P, Q, &c., do not contain y, but may containz. This form 
of the equation contains y only in the first power, and is therefore 
called a linear equation. Suppose, for illustration, (383) be of the 
third order, 


(384) d°y + Pd?ydx + Qdydz*? + Sydz* = o. 

Assume, 
(A) yeltt, «. dy=e“udz, Py = ef" (u'dz’ + dudz). 
(B) d’y = ef “**(wWdax*® + 3ududz® + d’udz). 


These values put into (334), it becomes, after dividing by dz, and 
by the exponential, 
(385) du + (3u + P)dudx + (wv + Pu? + Qu + S)dz* = 0, 
an equation of the second order in wu and z. 

If P, Q, and S be constant, we may take uw constant. For if u 
be constant, du = o, d’u = o, and (385) becomes, 
(386) Pt Pe + n+ Sb = 0, 
from which u may be determined. m 

Let uw’, u'’, u'’, be the three roots of (386). ‘Then designating 
the values of y in A, corresponding to these three roots by y’, y", 
y''’, we have, from the first of (A), 
C7): gS ey =e aa Ce 
or putting C’, C’, C’”, for e°, e°’, e°'”’, (387) becomes, 
(388) ae kre | oe ee oe Ge 

Each of these values satisfies (384), and are particular cases of 
the complete primitive of that equation. As the complete primitive 
contains each of the values (388), it may be represented by the sum 
of (388), viz: 
(389) y selene Ole Fe er, 

26 


282 INTEGRAL CALCULUS, 


For if we substitute into (384) the values of the differentials of 
(388), that equation will be satisfied. 

In the same manner may the equation (383) be integrated, when 
P, Q, &c., are constants. 

If P, Q, &c., are functions of x, then if we knew three values 
of y, viz: Cly', C"y", C’"y'’, which would satisfy (3884), the com- 
plete primitive would be their sum, viz: 

(390) y= Cy’ + C’y" + Ory". 

For if the several differentials of (390) be put into (384), then by 
putting (384) equal to M, and by putting M’,M"’, M’", for the cor- 
responding values of (384) in y', y'’, y'’’, we have, 

(391) C’M’ + C’M” + CM" = 0, 
a result in which M’, M”, and M'", are separately zero, since each 
of the values C'y’, C’y", C'"y'"’, satisfies (384), 

For more particular details on the integration of the form (383), 

consult La Croix, vol. II., p. 3183—3887. 


PROPOSITION XVIII. 


Determine the integral of a total differential equation of 
three variables, when the first side of the equation contains 
one variable, and the other side the other two. 

In the Differential Calculus, [Proposition LXIX.,- (377),], we 
found the total differential of the equation, 

(392) % = 9(#,y). 
to be, 
dzdz dzdy 
(393) dz = da + Sra 
in which each term on the second side is the partial differential 
of (392). 


R 


Case 1. 
Suppose now we have to integrate the equation, 
(394) dz = Pdx + Qdy, 


in which P and Q are any functions of x and y. 
If (394) has been derived immediately from differentiating an 


EQUATIONS OF THREE VARIABLES. 283 
equation such as (392), it is called an Exact Differential. The 
first examination of (394) is to ascertain if it be an exact differen- 
tial. If it be so, then P and Q are partial differential coefficients, 
and are equal to the partial differential coefficients of (393), that is, 

dz - dz 
395 P=‘ Bi fase eno 
as dx’ dy 
Differentiate the first of these relatively to y, and the second 
relatively to a, and we have, 


2a Q, 172 
(396) ae See Ohh Woes Sy 
dy dxdy dx dydx 
Since, [by Differential Calculus, Proposition LXVIII., Cor. —|, 
d*z a d*z we have, 
dxdy dydz 
(397) dP se a, 
dy dz 


When this condition exists, (394) is an exact differential. 

In the same manner, if we have an equation of four variables, 
(398) dz = Pdx + Qdy + Rdu, 
we have, in case it be on exact differential, the conditions, 
dP < dQ TA sae ileal eee dQ _ aR 


— —9) v 


dy dx’ du dx dz dy 
This we will exhibit at large hereafter. 
Ex.—Suppose we have, as a particular case of (394), the equation, 


(399) 


(a) dz = (6xy — y’)dx + (8x — 2ay)dy: Here, 

(b) P = 6ry — 7’, and Q = 32° — 2ry. 
Then differentiating, we have, 

(c hi = 627 — 2y, and te — 6x — 2y. 


dy dx 

These being equal, the condition (397) is fulfilled, and (a) is an 
exact differential. 

In order to integrate (394), when it is an exact differential of an 
equation (392), we know that the first term of (394) has been ob- 
tained by differentiating (392) for y constant. Hence if we integrate 
the first term, Pdx, supposing y to be constant, we will have, after 
the proper correction, the primitive (392). Hence the primitive 
of (394) is, 


a es ~~ an = as 
lina peas 3 


ait oath in Ae py REN 
: Se 


aetna ens- nied - 
— — ge aan a 


a 


284 INTEGRAL CALCULUS. 
(400) z=f Pde +C. 

But since y was made constant in differentiating for the first term 
of (394), C may contain y, and we may write (400) in the form, 
(401) z = f Pda + y', 
where y’ is put for the function of y and constants that disappeared 
in differentiating (392) for y constant. 

To determine y', differentiate (401) for y as the variable, and 
we have, 


(402) a te 


from which we have, 
' d : d { Pdx 
(403) 29. = q eae =f{/(2 Vay, 
dy dy 


This value put into (401), we have, for the complete integral 
of (394), 


404) Pd Q d { Pdx : 
( i ah D +f ( dy ) a 

If we had taken the last term, Qdy, of (394) and integrated, we 
would have obtained a form similar to (404), for the integral. 

To apply this process to example (a) above, we have, by inte- 
grating the first term of (a) for y constant, 


(d) z= 3da’y — yx + y’. 
This differentiated for y variable, is, 
dz dy' 
— == 3a? — 22 Eee |: But by (0), 
(e) apa Y Suite y (9) 
(f) 3x? — 2ry — Q. Hence from (403), 
dy 
o BL men fA or '—C. Hence (da) is, 
(g) dy y ( ) 
(h) % = 3a°y — yx + C, 


the complete primitive of (a). 
Ex. 2. Integrate the equation, 

(k) dz — ydx + (x + a)dy. 

Here P = y, and Q = a + a, and 


EQUATIONS OF THREE VARIABLES. 285 


Hence (k) is an exact differential. Integrate it for y constant, and 
we have, 


(2) = yxr+y’. 
Differentiate (1) for x constant, and we have, 
(m) saa = 2 - ay: —-Q=r+a, -. dy'=ay, -: y' =a, 
dy dy 
and (1) becomes, 
(7) S== yx + ay. 


Ex. 3. Integrate the equations, 
dz = 2xy — (x* + 2my)dy. 
Ans. % = x’y — my’. 
a ydx — «dy 
ar? rs y? x 
Ans. z = tan ~ +. C. 
y 


Case 2. 
If the second side of (394) be homogeneous as well as exact, the 
integral may be obtained in a very simple manner, Let 


(405) 2 = o(2Y), 
be the primitive of the homogeneous equation, 
(406) dz = Pdz + Qdy. 


Let n be the degree of homogeneity of (405), then nm — 1 is the 
degree of homogeneity of P and Q, in (406). 

Assume y = ra. This value of y put into (405), it becomes, 
(407) a= nae. 
where R is put for a function of r. These values of y and z put 
into (406), it becomes, [putting P’ and Q’ for what P and Q then 
become, | 
(408) d(R2”") = Pdz + Q'd (rz), 
or performing the differentiations denoted in (408), we have, 

(409) nRa”"dx + 2°dR = (P' + Q'r)dz + Q'adr. 

The first term on each side of this is the differential of (405) 
relatively to 2, and since these partial differentials must be equal, 
we have, 

(410) nRa™! — P’ + Q’r. 
U 26 * 


286 INTEGRAL CALCULUS, 


Restore in this the value of 7 from y = ra, and since P’ and Q’ 
then return to P and Q, (410) becomes, 


(411) nRae™ = P + qt, 
xr 
or eliminating R by (407), we have, 
(412) poiaaty Sy 
n 


the primitive of (406). Hence to integrate a homogeneous exact 
differential equation, we need only change dx and dy to x and y, 
and divide by the degree of homogeneity increased by unity. 
Ex.—Integrate the equation, 
(a) dz = 32°ydx + (a — 4y’)dy. 
This being both exact and homogeneous, the degree of homo- 
geneity being 3, its integral is, 
peak 32°y + (2° — 4y°)y 


(5) r = ay — y’. 
Ex. 2. Integrate the equation, 
(c) dz = (3ey? + y')dy + (yx — x)de, 
eal ary? ad x y? 
eT ny 


The constant should of course be added to complete the integral. 

The principle of this case is applicable in the same manner to an 
exact homogeneous equation of any number of variables. For 
example, if we have, 
(413) P dz = Pdx + Qdy + Radu, 
by an analogous procedure, we would get, [n — 1 being the degree 
of homogeneity of (413),] for the integral, 

_ Pe + Qy + Ru 


(414) z Ri At 
n 


PROPOSITION XIX. 


Determine the factor which will render a differential equa- 
tion exact. 
Take the equation of two variables. 


MAKING DIFFERENTIAL EQUATIONS EXACT. 287 


(415) Pdx + Qdy = 0, 
P and Q being functions of 2 and y. 

If (415) be an exact differential, it must fulfil the condition (397), 
and may be integrated by Proposition XVIII., without separating the 
variables. It may have happened, however, that a factor containing 
x or y, or both of them, has disappeared from (415), in consequence 
of which (415) is not an exact differential. If this factor could be 
found, and (415) multiplied by it, the equation would then be integ- 
rable as an exact differential. Suppose w be the factor that is 
required to render (415) an exact differential, w being a constant or 
a function of x or y, or of both of them. Multiply (415) by this fac- 
tor, and we have, 

(416) Pudx + Qudy = o. 

Since this by hypothesis is an exact differential, the condition 
(397) is fulfilled, and we have, 
dPu _ dQu 
dy da’ 
or performing the differentiation, we have, 
Pdu _ Qdu i dP _ dQ 

If this equation were integrated, we would have the value of w 
required. But (418) is more difficult to integrate in its general form 
than (415), and consequently the factor cannot always be deter- 
mined. ‘There are, however, two cases when (418) may be inte- 
grated, and the factor determined, 


(417) 


(418) u—o. 


Case 1. 
Let the factor wu be a function of 2 only. 


In this case, fe =o, and (418) becomes, 
ay 


dy dx7 Q u 
But since wis by hypothesis a function of 2, the first side of (419) 
cannot contain y. Hence we may assume, 


7p 
(420) ay (= a) Per 
Q \dy dz 


(419) ( __dQ\ dx _ dus 


and (419) becomes, 


os ae ae —- - pew mig 


ea 


Sas 


atin eee 
a 


SSE = =4 


oS 


ac ; 
aw 


Dee 


hte 
OM tas 
ee 
Wet J 
‘ey 
a ] 
nye | 
a a 


errs Ss 


288 INTEGRAL CALCULUS. 


(421) ie = fadz, +». logu= S Suede Soar, - i, aon eet 
u 


Then the equation (416) multiplied by e*’, is an exact differential. 
Ex. 1. Integrate the equation, 


(a) yor — xdy = 0. 
Here P = y, Q = — 2, and (420) becomes, 
1 2 
(b) afl baal oe 
—Zx —x 
and (421) gives, 
du 0: la , eon are 
(c) eT aap ages log.u = — 2log.x+log.c, -- u= = 


Hence (a) becomes, 
C (ydx — ady) —.0; whose integral is, Cy on Ci 
a eC 
Ex. 2. Integrate the equation, 
ady + (b — 2y)dx = 0. 


The factor is, wv =) +. 
x 


In like manner, if the factor were a function of y, its value may 
be obtained. 


Case 2, 

Let the equation (415) be homogeneous. 

Let u be the factor required, then (415) becomes, 
(422) Pudx + Qudy = 0 = dz. 

This equation being, by supposition, homogeneous, its integral is, 
by Case 2, last Proposition, [putting » — 1 for the degree of homo- 
geneity of (422),] 
(423) ee ae 

Divide (422) by (423), and we have, 

(424) Pudx + Qudy _ dz 
Pur + Quy nz 

The second side of this being an exact differential, the first side 
must also be an exact differential. Consequently (424) shows that 
the factor necessary to render (415) a complete integral, is, 


MAKING DIFFERENTIAL EQUATIONS BPXACT. 289 


1 
Ue eee —— « 
Px + Qy 
The same principle could be extended to a homogeneous equation 
of any number of variables. For if we had an equation as, 


(425) 


(426) Pdx + Qdy + Rdz= 0, 
the factor required is, 
(427) u t m 


~ Pe + Qy + Rz 
Ex.—Integrate the equation, 


(a) 2xydy — (y* — x*)dx = o. 
This being homogeneous, the factor (425) is, 
; 1 
(0) i sir 
ry” + 2 


Equation (a) being multiplied by this factor, may be readily put in 
the form, 


2ydy 4+ 2adxr _ (y? + 2°) dz 


(c) SA ma PES a 
y? 4 3 (y? + a”) az 
The integral of this is, 
(d) log.(2? + y?)—logcr = 0, *: 2+ y'—cxr=o. 


Besides these, an indefinite number of factors may be found, 
which will render integrable the equation (415). For if z be the 
integral of that equation, and w a factor which renders it exact, 


(428) dz = Pudx + Qudy. 
Multiply this by /z, and we have, 
(429) fudz = Pufzdz + Qufzdy. 


The form of fz being arbitrary, it may be taken of any value, 
as 2°, then 2*dz, the first side of (429), being an exact differential, 
the equation 
(430) wdz = wu (Pdx + Qdy), 
is also an exact differential, and the factor z°u renders (415) exact. 

The process of integration by first determining a factor, is, how- 
ever, in general complicated, and cannot indeed be practically ap- 
plied except in cases where the integral may be readily obtained by 
other processes, 


>| 


# 
A a 
gf. ten 
ig 
hs 
ee 
Ve 
ho 
are. 
i 
Lie be 
Pee ay 
Pharay 
i 
WD 
| 
th 
MI 
a 


es — 
en wn Dens 


i! 
hi 
ul 
i} 

* )4 
f 
ihe 
pay 
ea? 
Pes 
i 


290 INTEGRAL CALCULUS. 


PROPOSITION XX. 


Determine the integral of a differential equation of three 
variables, when the variables enter it in any manner. 

Let the equation be, 
(431) Pdx + Qdy + Rdz =0, 
where P, Q, R are each functions of 2, y, and z. It is evident that 
the variables in (481) cannot in general be separated so as to re- 
duce (481) to the form (394). But if the complete primitive 
of (431) be, 


(432) o(@,y;2) =-0 = 4, 
then since the differential of this is, 
dudx dudy dudz 
433 a, = 0, 
a dx ra dy Ry dz 
we have, by comparing (431) and (433), the conditions, 
du du du 
434 Diver at Gat Loa eS ee 
a) dx e dy dz 


Combining these two and two, as in (395) to (397), we have the 
conditions, 


(435) yi ge pt 
dy dx dz dx dz dy 
When these conditions exist, (431) is an exact differential. 
Suppose we have the equation, 
(a) 2zadx + 2’dy + (x? + 2zy)dz = o. 

Here P = 227, Q=2, R= a? + 2zy, and the conditions 
(435) are fulfilled, and equation (a) is an exact differential. 

If (431) does not fulfil the conditions (435), let us suppose (431) 
to become an exact differential, by being multiplied by some factor u. 
Suppose that 
(436) Pudx + Qudy + Rudz = 0, 
be an exact differential. Then the conditions (4385) are, 


(487) duP — duQ duP _ duR duQ _ duR 
pi, =< da} PARES We ares, 


EQUATIONS OF THREE VARIABLES. 291 


Performing the differentiations, these equations become, 


dQ Pdu R du 
8 oe me eee 
dy dx ) dz dx 


> Y k 
(438) 4 u (=. canoe =) -+- Pdu mone FY du — 0. 


dz dx dz dx 
uv (Se =) in UGE 2 RS = o 
dz dy dz dy 


Multiply the first of these by R, the second by — Q, and the 
third by P, and adding the products we have, after dividing by u, 


439) R Ch ae QS c+ abies pi Teese ee 
dy dx dx dz dz dy 

This is an equation of condition, without which (481) cannot be 
made an exact differential for any value of uw. 

Suppose we have, 
(d) (ay — bz)dx + (cz —ax)dy + (bx — cy)dz = 0. 

Here P= ay—bz, Q=cz—ar, R=bxr —cy, and 
(439) is satisfied, but (485) not. Hence equation (b) may be made 
an exact differential, by multiplying by some factor. 


Case 1. 

Integrate (481) when condition (439) exists. 

As (436) came from differentiating a function of a,y,z, first for 
one of the variables, asz constant, and then for each of the others 
tn succession as constant, we may, in returning to the integral, con- 
sider z as constant, which reduces (436) to 
{440) Pudx + Qudy = o. 

The factor u rendering this an exact differential, it is integrable. 
Let us denote its integral by V, and since the constant added may 
contain z, let us denote it by 2’, and the complete integral of (440) 
may be represented by 
(441) Vt2vk=o. 

Differentiate this for z, as the only variable, and we have, [since 
V is supposed to contain 2], 


(442) aVds ae a dade = 0, - Or, (2 4. sey dz =-0i 
dz dz dz dz 


292 INTEGRAL CALCULUS, 


As the coefficient of dz in this ought to be the same as the co- 
eficient of dz in (436), we have, by equating these coefficients, 


(443) Ru = ay +f first or 2' = f (Ru — oh dz. 
dz dz 


dz 
This value of z’, put into (441), gives for the complete integral 
of (431), the form, 


(444) Vv + f (Ru-L)a = . 


As 2' is, by supposition, a function of z, the value of z' in (443) 
can contain neither z nor y. If the conditions (435) are also ful- 
filled, then « = 1 in (440), 444), &e. 

Ex. 1. Integrate the equation, 

(ec) — vydx + vady + (2zry + 22?)dz = 0. 

This fulfils the condition (439), but not (434). Supposing z con- 
stant, (c) becomes, 

(d) 2 (xdy —ydx) = o. 

The part in the vinculum is an exact differential when the factor 
wis of the value, 


(e) “= = 
Multiplying by this factor, and integrating, we have for V, 
2 
Vas ye 
(f) : 


To determine z', observe that R = 2zry + 222 Multiply this 
by (e), and we have, 
(g) Ru = ates is 
Differentiate (f) for z variable, and we have, 
Ns 
HE er age, 
These values put into (443), we have, 
(h) a! = f 2dz = z+ C, 
Hence the complete integral of (c) is, 


2 
(k) ee we aC. 2g, 
64 


EQUATIONS OF THREE VARIABLES. 293 


The factor u required to render Pdr + Qdy = 0,a complete 
integral, may be found by the process pointed out in Proposition XIX. 
Ex. 2. Integrate the equation, 
(2) Qzrda + 2ydy + a’dz + adz = 0, 


(m) Anse 22+ y% + azr+ C= 0. 


Case 2. 

Integrate (481) when condition (439) does not exist. 

When (489) does not exist, we infer that no factor will render 
(481) an exact differential, and consequently there exists no single 
primitive whose differential is (431). On this account, when condi- 
tion (489) does not exist, (431) has been regarded as without mean- 
ing ; and indeed it is so, if an equation of three variables be re- 
garded as the equation of a surface. But Monge, and after him, 
writers on the Calculus, consider that when (439) does not exist, 
(431) may be satisfied by the two equations (441) and (443), in 
which x’ may be taken for any arbitrary function of z. These two 
equations, taken together, represent a curve of double curvature. 
Hence in this case, since 2 is entirely arbitrary, there are an indefi- 
nite number of curves of double curvature which satisfy equa- 
tion (431). 

As an example illustrative of this, take the equation 
(p) Qedx + 2ydy— dz = 0. 

This equation does not satisfy condition (439). Hence, to obtain 
an integral, take the part of (p) corresponding to (440), viz. 

(q) Q2dx + 2ydy. 

This being an exact integral, (441) becomes 
(7) 222 + yr + 2' = 0, 
and (443) becomes, (since R=—1), 


/ 
m dz 


(s) —l=2r + 


dz 
If z' be any given function of z, as, 
? C 
(t) z' = mz then __ = 2mz, 
az 


and the two equations (1) and (s) are, 
9» 
“<4 


294 INTEGRAL CALCULUS. 


22x 2 + mz? = o 
(2m) 22 ee of Lees B3 
which represent a curve of a double curvature. 
For any other value of z' we would have another curve, &c. 
This is the method of treating (431) when (439) does not exist. 


PROPOSITION XXI. 


Determine the integral of a partial differential equation 
of the first order containing three variables. 


A partial differential equation of the first order is one derived 
from an equation between three variables, and may contain the three 
variables, constants, and the two partial differential coefficients, 

dz dz 
en, Guu os 
dx dy 
which we will frequently denote by p and q. 
We will examine several cases of this proposition. 


Case 1. 
Let the equation contain but one partial differential coefficient. 
If this coefficient equal a constant, we have, 


(445) oe = 


The integral of this is, 
(446) | 2 = Mr + OY, 
where gy is added because (445) is supposed to be deduced from 
the primitive, on the supposition that y is constant. 
If the partial differential coefficient equal a function of x, we have, 
dz 
447 — = Or -- Z= x.dzx + oy. 
(447) souks J? ey 
If, for example, gx = ax? + bx, (447) becomes, 
‘rep AT oa 
le ie te 
st es ery: 
If the partial differential coefficient equal a function of y, we have, 


(448) —_— >= YY; 


PARTIAL DIFFERENTIAL EQUATIONS. 295 


and since this is to be integrated as if y were constant, the integral 
of (448) is 
(449) c= axuyy + oy 

If the partial differential coefficient equal a function of x and y, 


we have, 
(450) = = ¥ (x,y) a f ¥(ay)4e + oy. 


The integral of (450) is to be taken for y constant. 
Ex.—lIntegrate the equation, 


% 2 
(2) dz _ ar. g = a logy? + 2°) + oy. 
dx y+ x 
Ex. 2. Integrate the equation, 
1% br 2 2)" 
(b) Petes Se A aeeemg a 0 Gen @) er) PY 


dx (y? + ee 

If the variables x and z be not separated, we separate them by 
any of the methods applicable to the particular example. 

Illustrative of this, take the following example. 

Ex. 3. Integrate the equation, 
(°) das, edie 
Separate the variables x and z in this, and we have, 

dx _—s mdz 


x y? — az 


The integral of this is, 
m 


(d) log.2 = — — log.(y’ — az) + $Y- 
a 


Case 2. 
Let the equation contain two partial differential coefficients, and 
functions of # and y. 
We may represent such an equation by 
(451) Pp + Qq = 9, 
where P and Q are functions of x and y. 
Now since « is to be a function of x and y, we have the general 


form in Differential Calculus, (877), viz : 


296 INTEGRAL CALCULUS, 


(452) dz = pdx + qdy. 
Eliminate p between the two equations (451) and (452), and we have 
(453) dz = o (Pdy — Qdz). 


Suppose u the factor which renders Pdy — Qd@ an exact in- 
tegral, and put 


(454) u (Pdy — Qdx) = dV. 
This put into (453), we have, 
(455) Guin TEM 
Pu 


In order that this may be integrable, we must have aa a func- 
u 


tion of V, which we may represent by FV ; then (455) becomes, 

iM (456) dz = PVay; 

Nay The second side of this being a function of a single variable, is 
Ait integrable, and we have, 

ore (457) z= fFVdV = 9V. 


We will illustrate this by some examples. 
Ex. 1, Integrate the equation, 


see= 
ED Re pig rk tN 
> = < " 


a 
yrs - _ 
oe P 


Sy (a) px + qy =o. 
\ rs Here P = x, Q = y, and (454) becomes, 
kink (0) dV = u (ady — ydz). 


wi nena iene gem 
~ Oe 
= pir 


The factor u must obviously be u = a~*. Substituting this value 
into (b), and integrating, we have, 


(c) Vv= we 


HY 
This value of V put into (457), we have, 


@ Sea) 


for the complete integral of equation (a). 
Ex. 2. Integrate the equation, 


(€) py — qt =o 
Here P = y, Q = — 2, and (454) becomes, 
f) dV = u(adx + ydy). 


The factor u must here obviously be uv = 2. Put for w this value, 
and the integral of (f) is, 


PARTIAL DIFFERENTIAL EQUATIONS, 297 


(8) Vise eit y?. 
This value put into (45 Hebe ve have, 
(h) = G(x ma + y’), 


for the complete integral of (e). 

In this case, the integration depends upon the integrability of 
Pdy — Qdz ; and as this contains but two variables, x and y, this 
case depends upon Prop. XIII. : for we may put Pdy — Qdr = 0, 
and we have a differential equation of two variables. 


Case 3: 
Let the partial differential equation be of the form, 
(458) Pp + Qg@=R, 


in which P, Q, and R are functions of xz, y, and z. Writing as be- 
fore, the general form of a differential equation, viz : 

(459) dz = pdx + qdy, 

we eliminate p between the two equations (458) and (459), and we 
have, 

(460) Pdz — Rdx = q (Pdy — Qdz). 

Now there are two cases of (460), which may be noticed. 
First, the expressions Pdz — Rdz, and Pdy — Qdz, may contain 
only the variables x and y; or, Second, one or both of these ex- 
pressions may contain the three variables z,y,z. We will examine 
each of these cases. 

In the first case, we may find a factor wu which renders 
Pdz — Rdz integrable, and a factor u' which renders Pdy — Qdzx 
integrable. Hence assuming 
(461) u(Pdz — Rdz) = dV, and w' (Pdy— Qdzx) = dV’, 
(460) becomes, 

(462) dv = @ av’. 


w' 
This is not integrable unless 2" be a function of V’. But as 
wu! 
(459) is an exact differential, (462) must also be exact. Hence we 
u | 
may assume tae = FV’, and (462) becomes, 
U 


(463) dV = FVdV'. 


o~ «# 
27 


298 INTEGRAL CALCULUS. 


The integral of this 1s, 
(463a) V='oV', 
which is the compiete primitive of (458). 
We will illustrate this by an example. 
Ex.—Integrate the equation, 
(2) pr + qy = nz. 
Here P = x, Q = y, R = nz, and equations (461) become, 
(b) dV = tu («dz — nzdz), dV' = w' (ady — ydz). 
The factors which render these integrable are, 


“u= oy dat didpa Sumi) 
gt! x 
Multiplying equations (6) by these, and integrating, we have the 


equations, 


These values of V and V’ put into (463), we have for the com- 
plete primitive of equation (a), 


(d) z= 2.6 ( *), 


which shows z to be a homogeneous function of the other two 
variables. 

In the second case when 2, y and z enter into one or both sides 
of (460), we may still arrive at the integral by the following consi- 
derations, which embrace also the case just considered. 

In (459) p and q may be regarded as entirely arbitrary or inde- 
terminate coefficients: for that equation merely indicates that z is a 
function of x and y; and is the general form of the differential 
equation of three variables, whatever function z may be of a and y, 
Hence, q in (460) may be determined by the principle of the method 
of indeterminate coefficients. Hence, equating the coefficients. of 
the like powers of q, in (460), we have the equations, 

(464) Pdy — Qdz = 0, 
(465) Pdz — Rdz = o, 

If we were to eliminate q between (458) and (457), we would 

have, instead of (460), the equation, 


PARTIAL DIFFERENTIAL EQUATIONS. 299 


(466) Qdz — Rdy = p(Qdx — Pdy). 
For the same reason as before, we may put in this, 
(467) Qdz — Rdy = 09, 


and the factor of p in (466) gives (464). 

Hence, if we can integrate any two of the three equations (464), 
(465), (467), we can obtain the integral of (458). 

For these three equations of condition existing together, if we can 
integrate one of them as (467), then calling its integral, 
(468) F(z,y) = ©, 
we have, by solving it, the equation y = f (z,C), and this value of y 
may be substituted into (465). Equation (465) will then contain 
only 2,2, and the constant C, and may be integrated. 

Let its integral be represented by 
(469) ¥(2,y,C) + of = 9, 
where 9C is the constant added to complete the integral of (465). 
This constant ought obviously to be a function of the constant in 
(468) : for the two equations (467) and (465), existing together, the 
parameter, or arbitrary constant in one of them must be a function 
of the arbitrary constant in the other, Place the value of C from 
(468), into (469), and we have an equation containing «, y and =, 
the primitive of (458). 

4x. 1. Integrate the equation, 
(4) pe + qy = mx + vy? 

Here P = 2, Q= y, R = mx’ + y’),? and the three equa- 
tions of condition (464), (465,) (467), become respectively, 


(b) xdy — ydx = 0, 
1 

(c) rdz — m(x* + y*)* dx = a, 
1 

(d) ydz — m(x? + y’)* dy = 0. 


Equation (b) is the only one of these immediately integrable. 
Multiplying it by the factor u = 2~ and integrating, we have, 


(e) J = C- y= Oz. 
x 
This value of y put into equation (c), we have, 
} 


(f) da — m1 + CY? de = 0. s—m(1+C) x 4 9C=o. 
Put in this the value of the constant C in (e), and we have, 


300 INTEGRAL CALCULUS. 


(g) 2 — m(a2 + 92)? + (2). 


This is the complete primitive of equation (a). 

If two of the equations of condition are integrable, this process 
is very simple. As another illustration, take the following example. 

Ex, 2. Integrate the equation, 


(h) pre —qzy = y’. 
Here P = zz, Q = — zy, R = y’, and the equations (464), 
(465), and(467), are 
(k) zsady + zydxz = o. 
(2) gxdz + y’dxz = o, 
(m) zydz + ydy = o. 


Divide (k) by z, and (m) by y, and integrating, we have from 
(k) the equation. 


(7) xy = C, and from (m) the equation, 
(0) 2 oe y” = o.C. 

Put into (0) the value of C in (n), and we have, 
(P) 2 + y? = o(zy) 


for the complete primitive of (h). 

Ex. 3. Integrate the equation, 

— mzp + 2xrq = Ty. 

The integral is, y? — 2° = 9(2ay + 2’). 

If the three equations of conditions (464), (465), (467), contain 
each the three variables x, y, and z, we can only obtain the integral 
by differentiating the two equations (464), (465), and then one of 
the variables, as a and its two differentials, may be eliminated be- 
tween the four equations (464), (465), and their differential equa- 
tions. This will give an equation of the second order, between z 
and y. If this equation be twice integrated, we will have a relation 
between z and y, containing two arbitrary constants. By means of 
this relation, one of the other equations of condition may be in- 
tegrated, and the relation between «, y, and z, be obtained as before. 


PROPOSITION XXII. 
Determine the arbitrary function which enters the integral 
of a partial differential equation of the first order. 


PARTIAL DIFFERENTIAL EQUATIONS. 301 


As the determination of the arbitrary constant added in integra- 
tion, depends upon the nature of the particular problem which gave 
rise to the differential equation, so the determination of the arbitrary 
function which enters the integral of a partial differential equation, 
depends upon the nature of the particular problem which produced 
the equation, 

If this problem be restricted within definite limits, the arbitrary 
function which enters the integral may be determined by means of 
these limits. Suppose for example, we have the problem, 

Determine the surface whose section with ZY, is a given curve, 
and the trace of whose tangent plane on ZX makes with the azis 
of Xa given angle. 


FIG. 94. 


Put m for the tangent 
of the given angle DBX, 
and by the second condi- 
tion of the problem, we 
have the differential equa- 
tion, 

(470) io =" 

dx 
the integral of which is, 
(471) 2 = mz + oy. 

Now to get the section of this surface with the plane ZY, make 
x zero, and we have, from (471), 

(472) s = OY. 

And since the section is, by the first condition of the problem, a 
given curve, if its equation be, 


(473) 2 = Sy, 

then (472) and (473) being the same curve, we have, 
(474) oy = fy, 

and (471) becomes, 

(475) z2=mer + fy, 


where /y is known. 


Y 


302 INTEGRAL CALCULUS. 


{f, for example, the section on ZY were to be a circle, then (473) 

becomes, 
picts (R? ae y?)*, 
and the surface (475) is, 
z= mx + (R?— y*)’, 

We will give several Geometrical Applications, wherein the pro- 
cess of determining the arbitrary function will be elucidated. 

Definition.—Let us, for brevity, call the tangents of the angles 
DBX and DCY, [fig. 94], which the traces on ZX and ZY of the 
tangent plane to a surface make with the axis of X and Y, the tan- 
gents of X and Y respectively. 


APPLICATION 328th. 


Determine the surface whose section with the plane ZY 
iS a given curve, and in which the tangent of X varies as a 
given function of the co-ordinates # and y of the point of 
tangency. 

Recollecting that the partial differential coefficients p and q express 
the tangents of X and Y respectively, we have, by the second condi- 
tion of the Proposition, the equation, 


dz 
(476) == relay), 
Integrating this, we have, 
(477) z= ve o(v,y)dz + oy. 


If (x,y) = mzy, this becomes, 
fd 
(478) % == my a + 9y. 


If now the section of this surface, by the plane ZY, bea para- 
bola whose equation is, 
(479) 2 = py’, 
then (478) becomes, 
Cg : 
z= my oy + py’, 


which is the surface required. 


GEOMETRICAL APPLICATIONS. 303 


APPLICATION 39th. 

Determine the surface whose section with ZY is a given 
curve, and in which the tangents of X and Y are to each 
other in a given ratio. 

Let m be the ratio of the tangents, then by the Problem, we have, 


This is the differential equation, The integral of this is, by Pro- 
position XXI, Case 2, 


(482) ae 5 rata 4 y), or reciprocally, 
(483) RL 19) es. TF 

Let now the given section on zy be a parabola whose equation is, 
(484) y = az’. 


Since (488) gives for x = 0, y = 1%, we have 7% = az’, and 
(483) gives for the required surface, 
85) mz + y = az’, 
In this surface all sections parallel to the plane XY are straight 
lines, and all sections parallel to ZX or ZY are parabolas. 


APPLICATION 40th. 
Determine the surface whose section with ZY is a given 
curve, and in which the tangents of X and Y have the same 
ratio as the co-ordinates x and y of the point of tangency. 


By the Problem, we have, 


(486) Fie pecan dad ice p an ect 
q y dx dy 

The integral of this is by Proposition XXI, 
(487) i Se: as y'); oF or An 
(488) cae 


Suppose the section he by the stistabe on ZY be a parabola 
whose equation is, 
(489) a pz, 

This is the same section that would be given by making x = 0, 
in (488). For x = 0, (488) becomes y’ = xz. Equate this with 
the value of y? in (489), and we have, wz = pz. 


“i 
. 7 
} 
7 


304 INTEGRAL CALCULUS. 


Hence, (488) is, 

(490) e+ 4? = pr. 

This is the surface required, and is the paraboloid of revolution ; 
and, indeed, (488) is the general equation of surfaces of revolution, 
whose axis of revolution is the axis of Z, as is shown in books on 
Analytical Geometry. 

These applications exhibit the mode of determining the arbitrary 
function that enters the integral of a partial differential equation. 
If the problem that gave rise to the differential equation be not 
limited by a sufficient number of conditions, the integrated equation 
will contain the arbitrary function, without any limit for determin- 
ing it; but will still express some general truth in respect of all the 
cases that can be grouped under the given conditions which led to 
the differential equation. Thus, in Application 40th, if the condition 
of making a given section with the plane ZY were not given, we 
could obtain (488), but nothing farther. Now, if z be constant, 
(488) is a circle. Hence, a property common to all the surfaces 
that can be grouped under (488), is that all sections perpendicular 
to the axis of z are circles. For any particular surface of this 
description, as (490), we must have another condition, [as the first 
condition in Application 40th,] in order to give a definite form to the 
arbitrary function in (288). 


APPLICATION 4ist. 
Determine the surface whose section on the plane XY is a 


. given curve and whose tangent plane passes always through 


a given point. 

Take the given point on the axis of z at the distance m from the 
origin. The equation of the tangent plane is, 
(491) 2h — PoE) gy a), 
where x,y and z are the co-ordinates of the point of tangency. 

For the distance m we have from (491), 
(492) %—px—qy =m, or, px + qy=z—m. 

This is the differential equation. The integral of this is by 
Case 3, Proposition XXI, 


GEOMETRICAL APPLICATIONS, 805 


(493) ra 9(). 
x 4 


To determine the arbitrary function, we have recourse to the other 
condition of the proposition. Let the section on the plane XY be 
the hyperbola whose equation is, 

(494) ey =a 

This is the same curve that is given from (493), for z = 0. For 

z = 0 (493) becomes, 


(495) ao hee o(-“). 
4 & 


Since (494) and (495) are the same curve, let us determine the 
unknown function in (495). For this purpose, put the quantity in- 
volved in g equal an assumed value w, that is, assume, 


(496) J =u, 
2 
- Find the values of 2 and y from (494) and (496), and we have, 
(497) y= avu, and c= 
Vu 
These values put into (495), we have, 
ou = —m Ju 
a 


or restoring the value of wu from (496), 


(498) o(“) vy —- aN 


This value of the unknown function put into (498), we have for 
the surface required, 

(499) a’? (z—m)? = mary. 

This is the hyperbolic cone, whose equation was otherwise deter- 
mined in Differential Calculus, Proposition LXI. 

Without the condition that the surface have a given section on the 
plane of XY, we could deduce (493), which would appertain to all 
conic surfaces, whose vertex is on the axis of X, at the distance m 
from the origin, and which shows that in all such surfaces, the dis- 
tance z — m equals a homogeneous function of x and y. 

If the conic surface (493) had t curve of double curvature for its 


y 4 


306 INTEGRAL CALCULUS, 


directrix, the same plan would determine the arbitrary function. For 
let the surfaces whose intersection forms the curve of double curva- 
ture be 
(500) F (2,y,%) = 0, and F (24,2), 0 

Then assuming the terms involved in @ equal to w, as in (496), 
we eliminate x, y, and z from the four equations (493), (496) 
and (500), and we get, 
(501) gu = Fu, 
where Fu is some known combination of u. Restoring in (501) the 
value of u given in (496), we have a known function of the co- 
ordinates x and y, to put into (493). 


APPLICATION 42d. 

Determine the surface whose section on the plane of XY 
is a given curve, and whose tangent plane cuts from the axis 
of Za line whose length is in a given ratio to the abscissa 
y of the point of tangency. , 

Taking (491) for the tangent plane, and putting m for the given 
ratio, we have by the second condition of the problem, 


(502) %— px— qy = my. 
This integrated by Case 3, Proposition XXI., we have, 
(503) nite i logy = o(). 
my x 


If the section on the plane of XY be a parabola whose equation 
is, y = px’, (503) becomes, 


504 = my log._, 
( ) % my ih bere 


the surface required. 


PROPOSITION XXIII. 


Determine the integral of a partial differential equation of 
the second order. 


Let us, as in the Differential Calculus, assume the partial differ- 
ential coefficients, 


PARTIAL DIFFERENTIAL EQUATIONS. 307 


dz tz a*z d?z d*z 
— —Pp, a. te ———= 7 


= es 1 et es ey 
dx dy dx? dxdy ” dy? 
The most general form of a partial differential equation of the 


second order is, 
(505) O(L,Y,2,P,9,7,8,t) = o. 
We will integrate this equation in a few particular cases. 


(504a) 


Case 1. 
Let (505) contain but one of the partial differential coefficients, 
r, 8, or t, and constants. 
We then have, putting mm for the constant, 


ad’ 
506 ==. mM. 
2), an 
This integrated in respect of 2, is, 
(507) dz = Mr + oY, 
dx 


where @y is the constant added. 
Integrating (507) again, y being constant, we have, 

ss mx? 
(508) meee oes re 1a 
where .y is the constant added at the second integration. Equation 
(508) is the complete integral of (506). In like manner, if instead 
of (506), we had the equation, 

“ d’x 
(511) = mM, 

dx dy 

as this came from differentiating once for y variable, and once for 
x variable, we integrate once for y variable, by which we have, 


(512) ee = my + 92; 

dz 
and this we integrate again for x variable, by which we have, 
(513) 2= mary + i. ox dx + ¥y. 


Where ¥y is added to complete the integral. 


Case 2. 
Let (505) contain but one partial differential coefficient, and z 


and y. 


308 INTEGRAL CALCULUS. 


Solve the equation (505) in this case for the differential coeffi- 
cent, and we may represent the result by 
(514) Mek p 
dx? 
where P is put for a function of x and y. 
Integrate this twice for x as the variable, and we have, 


(515) z= f (/ Pax + oy) ae + vy. 


If the equation to integrate be, 

| (516) 28% ep, 

OR dx dy 

a This is to be integrated first for one of the co-ordinates, as x 


variable, and then for the other. Integrating for x variable, we 
have from (516), 


b 


a dzty” 
(517) FI ete. the 
Integrating this for y variable, we have, 
(518) i ff (Pdx + oy)dy + 92. 
Case 3. 


Let (505) be of the form 
dz dz 
519 aa a Wp See 
(619) da? a dx Re 
where P and Q are each functions of x and y. 
If we assume, 


a dz 
(520) 2 Ue 
Then differentiating, we have, 
dz. du 
dite. dx: 
and (519) becomes, 
(521) Wie. Q. 


dx 
This may be regarded as a partial differential equation in u and 


PARTIAL DIFFERENTIAL EQUATIONS. 3809 


x, and may be integrated for these as the only variables, a function 
of y being added to complete the integral. 

But since P and Q do not contain u, (521) is the same form as 
(205), whose integral is given at (212). Comparing (521) and 
(205), we may write the integral of (521), taking care to add in 
the integral (212), py for C. This integral in conjunction with 
(520) makes known the relation between 2,y and z, involved in 


(519). 


Case 4. 
Let (505) be of the form, 
(522) gre qioph ls? gy 
dx dy dx 


Differentiate the assumed equation (520) for y, and by substitution 
522) becomes, 
523) i Py = Cy. 
dy 
The same in form as (521), and integrated as the linear equa- 
tion (205). 
Case 5. 

Let (505) be in its general form. 

We may by examining it in this form, deduce a process for inte- 
grating it, in many particular cases. Suppose we have the equation 
(524) Rr + Ss'-+- Te = U, 
where R, S, T and U contain 2,y,z,p, and gq, in any manner. The 
general form of the first differential equation is, 

(526) dz = pdx + qdy, 
and as in Differential Calculus, at (399), we have, 
(527) dp = rdxz + sdy, and dq = sdx + tdy. 

We may eliminate ¢ and r between equations (527) and (524), 
and we have, 

(528) Rdpdy + Tdqdx — Udady = s(Rdy* — Sdady + Tdz*). 

Since the coefficients 7, s, and ¢ in (527), are indeterminate, we 
may equate the co-efficients of the like powers of s in (528), which 
.eads to the conditions, 

(529) <dpdy + Tdqdx — Udady = o. 


DR * 
Ke 


310 INTEGRAL CALCULUS. 


(530) Rdy*? — Sdardy + Tdz*? = o. 

Equations (529), (580), and (528) are analogous to (460), (464) 
and (465), and depend upon the same principle. 

As (530) contains dz and dy in the second power, we may solve 


it for “ and putting m and m’' for the roots, we have the two 
x 


equations, 
(531) dy — mdz = 0, and dy — m'dx = 0. 

These two values of dy substituted successively into (529), we 
have the two sets of equations, 

(532) dy — mdx = 0, Ridp + Tdq — Umdz = o. 
(533) dy — m'dz = 0, Rm'dp + Tdq — Um'dz = o. 

These equations exist in conjunction with (526). 

If we can integrate the equations (582), call their integrals V and 
V', then as at (460), (463), we have, 

(534) V= gV'. 

This is the first integral of (524), 

In like manner, if we could integrate equations (533), we would, 
by calling their integrals L and L’, have for the first integral 
of (524), 

(535) La @’L’. 

If now we integrate either of the equations (534) or (535), we 
will have the original function of (524). As the equations (534), 
(532) exist together, we may, if convenient, employ either of equa- 
tions (532) to aid in the integration of (534), 

If we can eliminate p and q between (534), (535), and (526), the 
result is an exact differential, because (526) is an exact differential, 
and the integral of this resulting equation may be obtained as in 
Proposition XVIII. 

As an example, take the equation, 

(b) r—at=o. 

Eliminating r and ¢ from this by means of (527), and equating 
the like powers of s, we get as particular forms of (529), (530), 
the equations, 

(c) dy? — a’dx* = 0, and dydp — a’dqdr = o. 


PARTIAL DIFFERENTIAL EQUATIONS. oll 


The equations (532), (538), become in this case, 


(d) dy — adx = 0, dp — adq = 9, and 
(e) dy + adx = 0, dp + adq = 0. 
The integrals of equations (d) are, since a is constant, 
(f) y—ar= Vi, and p—aq=V. 
Hence (534) is, 
(g) p— aq = Oy — a2). 
The integrals of equations (e) are, 
(h) y+ar=L' and p + aq:= L, 
and (535) becomes, 
(k) p+aqg=o(y + az). 


As we can here eliminate p and q between equations (526), (g); 
and (k), we have, after performing this elimination, the result, 
() d= o'(y + ax) (dy + adx) + 9(y — ax) (dy —adz) 

2a 

This equation falls under the process of Proposition XVIII, We 
may write its general form by observing, that since the integral of 
the form qudu, is some function of u, which may be represented by 
fu, so, putting u for y + ax, or y— az, in (1), we may write the 
integral of (2) in the form, 
(m) x = f(y + ar) + f' (y— a2), 
where the denominator 2a of (1), and the constant C, are included 
in the functions f and jf’. 

Equation (b) is the differential equation of vibrating cords, and 
occasioned, at one time, much discussion. 

As another example, integrate the equation, 
(n) xv’r + 2zrys + yt = 0. 

Proceeding as before, we find the values of m and m' to be equal, 
and equations (532) become, 


(0) ady — ydx = 0, xdp + ydq = 0. 
The integrals of these are, 
(2b) Y=—V', p+vVa=V, 
x 


consequently (534) becomes, 
(2c) p+ q Jj = (4). 
x x 


wok 


312 INTEGRAL CALCULUS, 


This is the first integral, and (2c) is to be integrated as equation 
(458), 
Comparing equation (2c) with (458), the equations (464) and 
(465), become, in this case, 
(2d) zdy —ydx =o, dz— o( 2) az oe Oe 
x 


The integral of the first of these is y — Cx, This value of y 
put into the second of (2d), the integral of that equation is, 
(2e) z— xg@C = ¥C, 
or restoring in this the value of C from y = Ca, we have for the 
complete integral of (n), 


a rant) 4(2) 


As another example, integrate the equation, 
(2g) qr — 2pqs + pt = o. 

Proceeding with this as in Example (n), we find the values of m 
and m’ to be equal, and the integration is performed as in Ex, (n). 

The result may be put in the form 
(2h) y = rox + Ys. 

As two arbitrary functions enter the integral of a partial differ. 
ential equation of the second order, we must, in order to determine 
these two functions subject the problem that produced the differential 
equation to two conditions besides the condition that produced the 
differential equation. For example, if we had the proposition, 

Determine the surface whose sections on the planes of ZY and 
XYare given curves, and whose second partial differential coeffi- 
cient at any point is constant, we could by the last condition form 
the differential equation of the Problem, viz. - 


dz 
536 Sa 
Cie oo 
The integral of this is, 
2 
(537) r= + oy + wy. 


Let the section on XY be a straight line, whose equation is 
y = ax, and the section on ZY be a parabola, z = by. ‘Then, 


when x = 0 (587) becomes z = .y, and since this is the same 
curve as 7 = by’, we have, 


PARTIAL DIFFERENTIAL EQUATIONS. 3138 


(538) dy = by’. 
Again when z = 0, (587) becomes, ma* + 2rpy + 24y = 0, 


which is the same as the straight line y = az, Eliminate x be- 
tween these two equations and we have 
(539) my? + 2ayey + 2a’yy = oO. 

The values of gy and yy from the two equations (538), (539) 
put into (537) we have for the particular surface required 
(540) 2az = ama? — xy (2a°b + m) + 2aby’. 

This will serve to show the method of determining the arbitrary 
functions that appear in the integrals of equations of the second 
order. 

PROPOSITION XXIV. 

Determine the integral of a partial differential equation 
of the first order containing four variables. 

If we have the equation, 

(541) z= $(2,y,u); 
we may, by putting p, q and s, for the partial differential coefficients, 
represent its differential by 


(542) dz = pdx + qdy + sdu. 
If now we have the partial differential equation, 
(543) Ns + Pp + Qq=R, 


in which N, P, Q, and R, are functions of a,y,z and u, we may 
eliminate one of the coefficients: as s between (543) and (542), and 
we have, 
(544) Ndz — Rdu = p(Ndx — Pdu) + q(Ndy — Qdu). 
Since p, g, and s, in (542), are indeterminate coefficients, p and q, 
may be regarded as indeterminate in (544), and in order to deter- 
mine them, we may put as in (460), the coefficients of p and q in 
(544), equal to zero, by which we have the three equations, 
(545) Ndz —Rdu =o, Ndx— Pdu=o, Ndy — Qdu=o. 
If these equations become integrable by being multiplied by the 
factors m, m', m'’ respectively, then by putting dV, dV’, dV”, for 
equations (545) respectively, and substituting into (544), we have 
the equation, 


314 INTEGRAL CALCULUS, 


pm dV’ 


Uy 


(546) dV = 


+ q —,aV". 
m 
This being an exact differential, we must have p as a function 
m 


of V’, and q ™ a function of V”. Consequently the integral of 
m" 


(546) is of the form, 

(547) V= ov’ + VV"; 

or since the two terms on the second side of (546) may be regarded 
as the two terms of a differential of a function of two variables, V' 
and V'’, we may, instead of (547), write the form, 

(548) = 9(V',V”). 

As the equations (545) exist together, we may obviously combine 
them in any manner with each other, and with the integrals of each 
other, which will enable us in general to integrate (548), when we 
can integrate any one of (545). 

The equations (545) may thus be treated in the same manner as 
the equations (464), (465), (467). 

Ex.—Integrate the equation, 


(a) us +ap+yq= 2%. 
Here N=u, P= #,Q = y, R= 2, and equations (545) 
become, 


(b) udx — xdu = 0, udz—2zdu=o0, udy — ydu= o. 
The integrals of these are, 


(c) Votre, ie anes 
Uu Uu Uu 


and (547) becomes, 


Oo Ene) +s(2) 


where any one of the variables is a homogeneous function of the 
other three. 


PROPOSITION XXV. 


Determine the integral of a partial differential equation 
of the first order, containing three variables, and exceeding 
the first degree. 


PARTIAL DIFFERENTIAL EQUATIONS. 315 


The general form for the partial differential equation of a function 
of three variables, is, 
(549) dz = pdx + qdy. 

The second side of this being an exact differential, must fulfil the 
condition of exact differentials at equation (397), viz: 
(550) id depot a 

dy dx 

But inasmuch as p and q, in (549), may contain z, we must dif- 
ferentiate p in regard to both y and 2, and q in regard to both « and 
z, This changes (550) to 


. if d d d 

551 v2 a9 — pit | D ae 

On) ae a Ge Eeink a 
If now, 

(552) P= '0, 


be a partial differential equation of three variables, containing p and 
q, involved to any power, as p’, q, pq, &c., then since (551) is the 
equation of condition to which (552) is subject, in order that (552) 
may be an exact differential, we eliminate p between (552) and 
(551), and we have a differential equation containing the four varia- 
bles x,y,z,g, and their partial differential coefficients in respect of q. 
The result of this elimination may be represented by 
(553) D ( xy.29 aq dq a == Os 
dz dy dz 

This equation may be integrated by Proposition XXIV., as an 
equation of four variables, and its integral being solved for q, we 
may represent it by 
(554) q = F (2,y,2,C), 
where C is the arbitrary constant introduced by integrating (553). 
From (554) and (552), we can eliminate q, and get the value of p, 
which we may represent by 


(555) p = § (2,y,2,C). 
These values of p and q put into (549), we have, 
(556) dz = F' (a,y,2,C)dx + F (2,y,z,C)dy. 


This being a total differential equation of three variables, may be 
integrated by Proposition XX. We may represent its integral by 


% — _ 
Sa — - ae 


ee es . 
a a te la I aN ee 


a 


Sn = als - get mT Oa =" i ar way _ os a 3 ae - Ea ee ee - . = . 
ee Se Ga be Si TS ee I ETS — —- aioe : % - 
- . ese : * tee = aoe Se ei es ee a a> atm a ; 
= = = > ee = aba : =~ en ee = : 
a ri ae A aes aad 


a 


316 INTEGRAL CALCULUS, 


(557) 1(23y32,C) = Ch 

The constant C’ may be a function of the constant C introduced 
in the integral of the equation of condition (558) ; but of this more 
hereafter. 

To reduce this process to a more practical form, suppose we 
differentiate (552) for 2,y,z,p and q, variables, and represent its 
differential by 
(558) dP = Adz + Bdy + Cdz + Ddp + Edgq = 0. 

Assume, 


(559) se =a"; a al pal 18 i =s', and 
Z y dz 
(560) dq = p'dx + q'dy + s'dz. 


Since p and q in (552) are functions of x, y and z, we must, to 
obtain the partial differential coefficients, 


in (551), first put 2 and a constant in (552), and then @ and y con- 
stant. This is the same as putting first dr = 0, and dz = 0, in 
(558), by which we have, from (558), 


(561) GP isos Baal: Bates 
dy D 
and next, putting in (558), dz = 0, dy = 0, by which we have, 
(562) Spal i Gaikl- Bigs 
dz D 
The values (561) and (562), put into (551), we have, 
(563) Dp’ + s'(Dp + Eq) + Eq’ = — (B + Cq). 


Eliminate p’ between (563) and (560), and putting the coefficients 
of s' and q’ in the resulting equation equal to zero [according to 
Proposition XXIV, (545)], we have the equations, 

( Ddy — Edzr = o, 
(564) 4 Ddz — (Eq + Dp) dz = o, 
. | Ddq + (B + Cq) dz = o. 

From these equations we can eliminate p by means of (552), and 
equations (564) may then be integrated as (545) in last Proposition. 

The equations (564) exist in conjunction with (549;) and, indeed, 
if we eliminate E between the two first of (564), we get (549). 


PARTIAL DIFFERENTIAL EQUATIONS, ol7 


It frequently happens that the last of (564) is immediately in- 
tegrable, in which case the other two are not needed, 

We will give a few examples illustrative of this theory. 

Ex. 1. Integrate the equation, 


(a) xp — aq? —2z= 0. 
Differentiate this, and comparing the differential with (558), we 
find A=p, B=o, C= —1, D=2, .E = 2aq, and 


equations (564) become, 
(b) «dy + 2aqdxr = 0, rdz + (2a + 1) qdz = 0, adq — qdx = 0. 
The last of these is immediately integrable, and gives us, 
(c) giant Oa. 
By means of equations (a) and (c), eliminate p and q from (549), 
and we have, 


(d) dz = (= a abiz) dx + Cady, 
3 


which is to be integrated by Proposition XX. Peérforming this inte- 
gration, we will have the required relation between 2, y and z. 

We will now examine several cases in which equations (564) can 
be integrated. 

If we put the last of equations (564) in the form 


3 ¢ 
(565) dq + (- + se = 0, 


it is obvious that it will be immediately integrable when the pro- 
posed equation is such that the coefficient of dx in (565) contains 
only g and x. In other words, that (565) may be immediately in- 
tegrable, we must have, 
(566) > + <4 = 9a) 
where @ denotes any given function, 

Whenever (566) exists, the value of q may be obtained imme- 
diately by integrating (565.) 

Again, if we eliminate dz between the first and last of (564), 
we have, 


nen B Cq 
567 lq + (= ) a Stig, 
(567) dq ast ad 

29 


he er 


Bh) 2 ee 


i fy Re ee RRO as aes 
A parr el a eg Semana —_—-* cae Oy Sneed 2 
= = ag one = - . _— ae lee > Sa - ee ona ~_ wee > ~ ~ 
oe me a = (is, Fane Sy TT a 


= ee Ses ee 
Sas et - ; 
ARIES 


318 INTEGRAL CALCULUS. 


which is immediately integrable if the proposed equation (552) be 


such that we have, 
Bay 

567a — att = e 
(567) ee 

Whenever this equation (567a) exists, the value of q may be ob- 
tained immediately by integrating (567). 

Again, if we eliminate dx between the two last of (564), we have, 

B + Cq 

568 dq + laa) aso, 
Ce) Dp + Kg 
which is immediately integrable if the proposed equation (502) be 


such that we have, 
B + Cq 


569) eee ,%). ! 
: Dp + Eq i 7 
Ex. 2. Integrate the equation, 
(e) p—qmtigrt+z=o. 
This leads to the condition, ; 
(f) dq =0 - g= C, 


and (e) becomes immediately integrable. 
As another example, take the equation, 
g) p—ye + qe az = 0. 
This gives for the last of (564), 
(h) dqg— (¢ + «)dx = 0, 
which is immediately integrable. 
Ex. 3. Integrate the equation, 


(k) ax —q@y—pr+z=o. 
For this example, (567a) exists, and (567) becomes, 
dy 
l Ai tae LO, 
(1) q—(l—4) saline 


which is immediately integrable. 

Besides the cases when one of the equations (566), (567a), and 
(569) exist, we may frequently integrate two of the equations (564) 
together, when these two contain between them but three of the four 
variables, q, x, y, and z, The method of integrating in such a case 
is pointed out at the end of Proposition XXI., for the equations 
(464), (465), and leads generally to an equation of the second order. 
Another plan is to integrate the three equations (564) together. 


PARTIAL DIFFERENTIAL EQUATIONS. 319 


Again, if we assume, 


(570) v= px + qy—%, 
we have an extensive class of partial differential equations of the form, 
(571) (2,Y,P.g,0) = 0 

[f we differentiate (570), we have, by virtue of the equation, 
(572) z= px + qy, the result, 
(573) dv = «dp + ydq. 


If we regard p and q as the independent variables, x and y as 
the differential coefficients, and vas the function sought, we may 
integrate (571) and (578) as we integrated in Proposition XXI. ; 
that is, eliminate one of the co-ordinates, as z, between (571) and 
(573), and from the resulting equation form two equations of condi- 
tion, by equating the coefficients of the like powers of y, which may 
be integrated as we integrated (464), (465). If we call the in- 
tegrals of these two equations of condition m and n, we may elimi- 
nate p and q between the equations m,n, and (571). The result 
will be the integral required, and will contain two arbitrary constants. 

Illustrative of the plan of integrating (571), take the following 
example. 

Ex.—Integrate the equation, 


(2a) pe + gy —v=o. 
Eliminate x between (2a) and (573), and we have, 
(2b) pdv — vdp = y (pdq— qadp). 


Zquating the coefficients of the like powers of y in (2b), we have 
the equations of condition 
(2c) pdv — vdp = a, pdq — gdp = 0, 
whose integrals are, 


(6 v 
(2d) hats |. log.p + == 1, 


1 
Pp q 
Eliminate p, g, and v between (2d), (2a), and (570), the result 1s 
the integral of (2a). 
If the proposed equation be of the form 
(574) v =F (p,q), 
then since this does not enable us to eliminate x or y from (573), we 
may put the coefficients of x and y in (573) separately zero, by 
which we have the equations, 


320 INTEGRAL CALCULUS. 


(575) dp=o, dq=0, -- p=m, and q= 2, 
and (574) becomes, 
(576) 2— mx —ny = F (m,n). 


Hence we see that the form (574) is integrated by putting the 
constants m and n for pand q, in (574). 

The relation between the constants m and n, and their determina- 
tion, we will point out in the next Proposition. 

The integration of partial differential equations may be frequently 
simplified by introducing an indeterminate coefficient. Suppose the 
equation be, 

(577) F (p,«) = F’ (qy)- 

By putting F (p,z) = #, we have also F" (q,y) = «, and these 
two equations being solved for p and q, we have p = f (a,), and 
q =f' (y:?)- These values of p and q put into (549), we have, 
(578) dz =f (#,a)dx + f' (y,w)dy. 

If « can be taken as constant, this may be integrated, and we have, 
(579) a= ff (to)dx + ff (ysw)dy + ©, 
where C may be a function of «. 

We will now give some Geometrical Applications involving differ- 
ential equations of the higher degrees. 


APPLICATION 43d. 
Determine the surface whose tangent plane passes through 


a given point, and cuts off, with the co-ordinate planes, a 
FIG, 95, 


given pyramid. 

Let the given point be 
on the axis of Z. 

Let the equation of the 
tangent plane be 
z'—2=p (x'—2) + 4(y'—y) 
where 2’,y',z', are the co- 
ordinates of any point on 
the plane. 

For the intercepts of this 
plane with the axes, we 
have, 


GEOMETRICAL APPLICATIONS. 321 


(580) AB=— rn he me a AD=z—pxr—qy- 
P q 
Put b° for the solid content of the given pyramid, and a for the 
distance AD, and by the problem we get the equation 


3 2 
(580a) 2 — pr—qy = ( 6b* pq ) 
a 
This being of the form (574), the complete primitive is, 
23 2 
(581) z— mre —ny = (= m ") 
a 


We will examine this result further in the next Proposition. 


APPLICATION 44th. 


Determine the surface whose tangent plane cuts off, with 
the co-ordinate planes, a given pyramid. 

Putting s* for six times the volume of the pyramid, and using the 
intercepts (580), we have, by the problem, 


: 1 
(581) z— px — qy = 8 (pq)*, 
which being of the form (574), its integral is, 
1 
(582) 2— mr — ny = s(mn)s, 
We will examine this further in the next Proposition. 


APPLICATION 45th. 


Determine the surface whose tangent plane cuts off, on the 
plane of XY, a triangle whose area varies as the square of 
the sum of the co-ordinates x and y of the point of tangency. 

Taking the notation (580), and putting for the area, e (x + y)’, 
we have, by the problem, é 


(583) z— pe —qy = (x + y) (2epq)*, 
where e is any constant. As this is of the form (571), then by 
eliminating a between (583) and (573), and equating in the result- 
ing equation the coefficients of the like powers of y, we have the 
equations, 

(584) —dp+ dq = 09, (2epq)" dv — vdp = 9, 


whose integrals are, 36 


) 
We 


a 


322 INTEGRAL CALCULUS. 
(585) q—p =m, log. . = Tt. 
(2e)**" (p*+ pm)’ 


Perform the integration of the last of (585), and then eliminate 
Pp, q and v between (585), (583), (570), and we have the primitive 
of (583), which makes known the surface required. 


APPLICATION 46th. 


Determine the surface which cuts a given family of sur- 
faces at a given angle. 

This is analogous to Application 26th. By a family of surfaces 
is understood all the surfaces that can be expressed by a single 
equation, when different values are given to a parameter in that 
equation. Thus the family of paraboloids of revolution may be ex- 
pressed by the equation, 

(586) m+ yy = pz, 
which, when @ takes different values, gives the family of paraboloids 
of revolution. 

Let the equation of the surfaces cut be represented by 


(587) Fife'iy' 2, By e200, 
and let the surface required be, 
(588) f (a sy!"2") = 0. 


The angle of two intersecting surfaces is the same as the angle 
formed by their tangent planes at any point of their common 
section. Let 


(589) s—s' = p(rt#—a2')+q(y—y), 
be the tangent plane to (587), and 
(590) 2—2' =p («—2")+q' (y—y") 


be the tangent plane to (588). 
af m be the angle which these planes make with each other, we 

have, by Analytical Geometry, 
aan 4 abate’ 


(591) | Cosim = a. 


1 
(Fa a ad a aay Herta a P) 
Supply in this the coefficients p and q from the given surface 
(587), and then eliminating B between (587) and (591), we have 
the differential of the surface required. 


GEOMETRICAL APPLICATIONS. 323 


If the angle m be a right angle, then (591) becomes, 
(592) 1 + pp’ + qq = 9, 
and the integration is effected by Proposition XXI. 

As an example of the integration find the surface which cuts at a 
given angle all the cylinders expressible by the equation, 


(a) a? oi grt. 
Here p = 0, q = 0, and (591) becomes, 

} v 1 

(5) Cos.m = — 


a Veg a ie SC 
From this we get, 
(c) Fa ada 


1 — cos.?m 


“NO 
“ 


where we put, for brevity, C* for 


cos.” 

The process pointed out at (577) will apply here ; for if we put 
p' = Ccos.m and q' = C sinm, equation (c) is satisfied. Hence, 
(578) becomes, 
(d) dz = Ccos.m dx + C sin.m.dy, 
which since m may be taken constant, gives for the integral, 
(e) 2 = Cr cos.m + Cy sin. m + Nn. 

The same result could be obtained from (564) by observing, that 
in this case, the last of (564) gives dq’ =o ~- q = a constant, 
which put into equation (c), gives p’ = a constant. 


The integrations in the following examples, under this Applica- 
tion, are performed by Proposition XXI. 

Ex.2. Find the surface which cuts at right angles all the family 
of paraboloids whose equation is, 


(f) x? + y? = 4az. 
\ ¢ 9 ] 
(gz) Ans, 227 + 2? + y? = 2 (2) , 
Ex. 8. Find the surface which cuts at a right angle the spheres 
whose equation is, 
(h) a ebay tea gts R* 
Zz 2 
(k) aneets — o(), 
7 x 


any conic surface whose vertex is at the centre. 


324 INTEGRAL CALCULUS, 


Ex. 4. Find the surface which cuts at a right angle all the 
spheres which touch a given plane at a given point. 

Let the plane XY be the given plane. ‘Take the given point for 
the origin, and for the family of spheres we have the equation. 


(2) ao? + y? + 2? = 2Rz, 

and for the trajectory required, we get, 

(m) a? ty? + 2% = zo(“). 
z 


APPLICATION 47th. 


Given the equation of a surface, determine the equation 
of an equivalent surface. 


Definition. 
Equivalent surfaces are those on which if equivalent areas be 
taken, the projections of these areas on a given plane will be equal. 
Let the plane of XY be the plane of projection; and let 


(593) % = (x,y), 
be the given surface; and suppose that 
(594) z= (x,y); 


be the required equivalent surface. 

Putting p' and q’ for the differential coefficients of (593), and 
p and q for the differential coefficients of (594), and equating the 
areas of the equivalent surfaces as given by (167), we have, 


u! 1 

(595) ff dady(1 + p? + 9”)? = f fdxdy(l + p+ q)’. 

Since the projections on ay of equivalent areas of the surfaces 
are equal, the dady of (595), is the same for both surfaces. Hence, 
omitting the signs of integration, and dividing by dy da, (595) 
becomes, 
(596) Didige ss p +a 

The values of p' and g' can be supplied in terms of x and y, from 
the given surface (593), and (596) will give for determining the 
equivalent surface, 


(597) p+ gi = F(a,y)- 


PARTIAL DIFFERENTIAL EQUATIONS. 325 


The integral of this will give the required surface (594). 

Ex. 1. Let the given surface (593), be the paraboloid of re- 
volution. 

Here (593) becomes, 
(a) 4mz = vt iy’, 
and equation (597) becomes, 
(5) Pa ea 
which, when integrated, will be the equivalent surface of (a). The 
integration belongs to Proposition XXV. 

Ex. 2. Let (598), the given surface, be a plane. 

Taking for the equation of the given plane, 


(c) z— me +ny + B, 
(597) becomes, 
(d) P+ Ge = (w+ ny = C, 


a constant. This is integrated as equation (c), in Application 46th, 
and gives a plane. 

It is obvious, from the definition of equivalent surfaces, that if a 
right cylinder be erected on the plane XY, it will cut equal areas 
from the two surfaces. 


PROPOSITION XXVI. 


Determine the singular solution of a partial differential 
equation of the first order. 

In Proposition XV., we observed that the singular solution of a 
differential equation of two variables depended upon the principle for 
determining the Locus of the intersection of consecutive lines. A 
similar remark applies to differential equations of three variables. 
In the preceding proposition we have seen, that in determining the 
complete primitive of a partial differential equation exceeding the 
first degree, two constants were introduced, as at (554) to (557), or 
at (570) to (576). Suppose m and n be these constants. The 
conditions of the problem that gave rise to the partial differential 
equation may be such that we can determine one of these constants 


326 INTEGRAL CALCULUS, 


in terms of the other, so that, putting m = on, the integrated equa- 
tion instead of being of the form, 


(598) F(z,y,z%,m,n) = 0, 
will be of the form, 
(599) F(x,y,2,n,gn) = 0. 


The conditions of the problem may, however, be such that no 
relation can be established between m and n; in which case they 
will be independent of each other. If the relation m = on, can 
be established from the conditions of the problem, the parameter n 
may enter into (599) in such a manner that we can eliminate it by 
the principles of consecutive surfaces, as detailed in the Differential 
Calculus, Proposition LIX. ; that is, we may eliminate n between 
(599) and its differential for m variable. ‘The result of this elimi- 
nation will be a surface which may be represented by 
(600) o(@,y,%) = 90. 

This surface, as is shown in the Differential Calculus, Proposition 
LIX., Cor, 3d, envelopes all the surfaces (599). ‘The result (600) 
is called the singular solution of (599). 

If the parameters m and n are independent of each other, we 
may differentiate (598) first for m variable, then for n variable, and 
we have the equations, 

(601) CEE ae a NM ie ts 
dm dn 
where, for brevity, F is put for F (2,y,z). 

If m and n be eliminated between (598) and (601), the result 
will be an equation of the form (600), which will be the envelope of 
all the surfaces comprised in (598). 

The result (600) is in this case called the singular solution of 
(598). 

We will illustrate this by a few particular examples. 

For this purpose, resume Application 48d, which proposed to 
Determine the surface whose tangent plane passes through a given 
point on the axis of Z, and cuts off, with the co-ordinate pianes, a 
given pyramid. 

The conditions of the problem gave for the integral of the differ- 
ential (580a),the equation, 


PARTIAL DIFFERENTIAL EQUATIONS. 327 


6b°nn \2 
(602) z— mx — ny= ea ; 
a 
Now the condition of passing through a given point on the axis 


of Z gives us, [since in (581), p = m, and gq = nj, the relation, 


603 zZ— mr — ny = a. 
( ; y 
Hence, 
6b°>mn \e a’ 
(604) = (- ) PO =: ON. 
a 6b'n 


This is the gn which we are to substitute into (602) for m. 
Making this substitution (602) becomes, 
eon a a ie Sate 
(605) z ne x ny = 

Differentiating this for n as the only variable, we have, 

(806) OF _y=o. 
6b°n? 

Eliminate n between (605) and (606), and we have, 
(607) 3b°(z — a)? = 2a° xy. 

This is the singular solution of (602). It is obvious that the 
process here is precisely that given at Proposition LXI, Differential 
Calculus, where the same problem is solved, and the result (607) 
obtained. 

The surface (607) is a single surface, while (602) the complete 
primitive is a family of planes which are all touched by (607). If, 
instead of taking the point through which the tangent plane passes 
to be on the axis of z, we had taken its co-ordinates (a',b’,c’,) the 
solution would be obtained in the same manner. As an illustration 
of the case where no relation can be fixed between m and n by the 
conditions of the problem that produced the differential equation, 
resume Application 44th, which proposed to Determine the surface 
whose tangent plane cuts off with the co-ordinate planes a given 
pyramid. The conditions of the problem gave a differential equa- 
tion whose integral (582), is 1 
(608) %— mx — ny = s(m n)°. 

Here there is no condition of the problem which enables us to fix 
any relation between m and n, ‘These parameters, therefore, are 
independent of each other. Hence, differentiating (608) first for m 
variable, and then for n variable, we have for (601), the equations, 


328 INTEGRAL CALCULUS. 


Abs 


cobs 


s 
ndm, — ydn = (=) mdn. 
vu 


© 


(609) — xdm = =(m) 
Eliminate m and n between the three equations, (608) and (609), 
and we have, 


(610) Lye = 


3 


% 


~) 


9 

This is the singular solution of (608), and is the surface to which 
all the surfaces (608), are tangent. 

This singular solution is the Proposition LXIII., Differential Cal- 
culus, where the same result is obtained. 

If the quantities m and both disappear when the equation is 
differentiated for them as variables, we conclude such a primitive 
does not admit of a singular solution. 

If the conditions of the problem which produced the complete 
primitive do not enable us to fix a relation between m and n, it might 
occur that one of them, as , would disappear from (601), and the 
other, m, remain. In this case we can eliminate m from the com- 
plete primitive, and the result will be an equation which will satisfy 
the proposed differential equation. 

As an example, resume Application 46th. ‘The first example 
under that application proposed to Determine a surface which would 
intersect at a given angle a family of cylinders, Equation (c) of 
that application is, 

(611) De Ange sae 

This is integrated at (e) in that Application, or we may integrate 

it in the usual way by equations (564), which leads tog = m,a 


constant; hence, p = (c? — m’)*. 
hese values of p and q put into the equation dz = pdx + qdy, 
we have, by integrating, and adding the constant n, 
(612) 2= (C— m)\? x + my +n. 
As the problem furnished no condition for establishing a relation 
between m and n, we may consider them independent. If we differ- 
entiate (612) for m variable, we get, 


(613) Eee a eae 


(C— m?\* 


PARTIAL DIFFERENTIAL EQUATIONS, 329 


The differential of it for n, leads to no result, as m disappears :n 
the differential equation. 

‘liminate m between (612) and (613), and we get, 

(614) (2—n) = c(a? + y°), 

the equation of a cone with its vertex on Z, and the axis of Z for its 
axis. The vertical angle of the cone depends on the given quantity 
c, which is a function of the given angle of intersection of the sur- 
faces. Equation (614) contains the arbitrary constant n which 
determines the position of the vertex. We would obtain the same 
result by differentiating, for m variable, the integral (e), of Applica- 
tion 46th, and eliminating m from (e), and its differential. 

As (614) contains but one arbitrary constant 7, it may be regarded 
as the general integral of (611). ‘The values of p and q deduced 
by differentiating (612) satisfy (611); and the values of p and q 
deduced from (614), and put into (611), also satisfy (611). 

The relation between the surfaces (612) and (614), is obvious : 
for if n be made constant, and m be left arbitrary or variable, (612) 
is a system of planes passing through a point on the axis of 2, each 
plane making a constant angle with that axis; and for the same 
value of n (614) is a conic surface, to which all these planes are 
tangent. Hence, for the same value of n, (614) is the singular 
solution, and (612) the complete primitive of (611). But as m is 
arbitrary, (612) denotes an infinite number of such systems of 
planes, and (614) an infinite number of such cones ; each cone 
being the assemblage of the characteristics of the system of planes 
that pass through its vertex. ‘There are, therefore, an infinite num- 
ber of such systems of planes, each plane of which satisfies the 
problem ; and to each system of planes there is a cone tangent to 
all the planes of the system, which cone also satisfies the problem. 

Equation (612) is, therefore, the most universal solution the 
problem admits of; and (614), though a singular solution in respect 
of any one of the system of planes comprised in (612), is, never- 
theless, a general solution, inasmuch as it contains an arbitrary 
constant 7. 

If the problem proposed were, “A plane passes through a given 


point ona fixed line and makes a given angle with the line, deter- 
30 


330 INTEGRAL CALCULUS. 


mine the surface to which the plane is tangent,” we would by 
taking the fixed line for the axis of Z find (612) for the equation of 
the plane where n is the distance on the axis of Zof the given point 
from the origin, Eliminating m from this plane by the principle of 
consecutive surfaces given in the Differential Calculus, we have (614). 
As the equation (611) is independent of the position of the axis of the 
cylinder, and is the same whether the cylinder be 2? + y? = R?’, 
or (x — a)’ + (y — 6) = R’, we infer that the systems of sur- 
faces (612), (614), may have any line parallel to the axis of the 
cylinder, for the line of vertices of the cone, or line of intersections 
of the systems of planes. Hence, any point in space may be taken 
as the point through which the system of planes (612) passes. We 
may express this analytically if we denote by (a,b,e,) the co-ordin- 
ates of such a point in space; then for this point (612) becomes, 


ai 
(615) e= (Cc —m)* a + mb +a, 
and subtracting this from (612), we have, 


(616) z—e= (c2 — m2)? (c —a) + m(y — 5). 

If we eliminate m between (616), and its differential for m 
variable, we have, 

(617) z2—e=cJ/(«—a)y+ (y— by, 
which is a cone with its base parallel to the plane zy. 

The same remarks apply to the relation between (616), (617), as 
were made concerning (612), (614). 

It is obvious that (615) furnishes a relation between m and n, in 
terms, however, of three arbitrary constants a,b,e. If a,b,e, be 
considered as known, we have simply another condition added to 
the problem, viz. that the surface which intersects the family of 
cylinders shall pass through a given point, (a,b,e). 

If now we had two of the constants a,b,e, in functions of the 
third, [which we would have if the point (a,b,e,) were on a curve 
of double curvature,] we might eliminate this constant and obtain 
the singular solution under this restriction. But, as there may be 
an infinite number of curves of double curvature on which the point 
(a,b,e,) may be situated, there would also be an infinite number of 
singular solutions of this. description. 


PARTIAL DIFFERENTIAL EQUATIONS. 331 


If any one of the constants a,b,e, be zero, and the other tws 
related as in a plane curve, we can determine the singular solution 
under this restriction. 

This discussion of (612), (614), and (616), (617), exhibits the 
mode of determining the constants introduced into the integrals of 
partial differential equations of the first order exceeding the first 
degree. The nature and conditions of the problem that produced 
the differential equation are the guide. 

The same discussion and remarks apply to Application 47th, Ex- 
ample 2, which proposed to find a surface equivalent to a given 
plane. 

As another example of singular solutions, take the following 
problems. 


APPLICATION 48th. 

Through a given point a plane is drawn, cutting off on 
two of the co-ordinate planes triangles, the sum of whose 
areas is constant, determine the surface to which this plane 
is tangent. 


APPLICATION 49th. 
A plane is drawn tangent to a surface, its traces on the 
co-ordinate planes form a triangle whose area is given, de- 
termine the surface. 


APPLICATION 50th. 
A plane is drawn tangent to a surface, its traces form 
with the co-ordinate axes triangles, the sum of whose areas 
is given, determine the surface. 


CLL LDP LDL LL ALD PA OL DL LE 


re 


332 INTEGRAL CALCULUS. 


MISCELLANEOUS PROBLEMS. 


PROBLEM A. 


A curve is traced on a given cylinder, determine the curve 
when the cylinder is developed on a plane. 

Call the right line which generates the cylinder, the element of 
the cylinder. 

Let (2’,y',z',) be the co-ordinates of any point in space, then for 
the element of the cylinder, we have the equations, 
(1) xi —x=a(z'—z), andy’ —y= b(z' —2z). 

Let the given curve traced on the cylinder be represented by 
(2) LT = wz, Y= V2. 

Let the tangent line to this curve be, 
(3) r—x2 = p(z'—z), and y' —y = q(z' —32). 

Let P be the angle which the tangent line (3) makes with the 
element (1), and we have, by Analytical Geometry, 

1 + ap + bq 1 

(4), -Cos.P. = i, Where M = (1 + a? + 6*)?° 

M(1 + p? + q’)° 


FIG. 96 


Suppose now the cylinder be de- 
veloped, and that OR be the develop- 
ment of the curve. The elements of 
the cylinder are the ordinates of OR. 
Let OV, one of these elements, be 
the axis of V, and call the axis of Oo 
abscissas the axis of U. It is obvious that the angle P will be the 
same before and after development. Hence, if HP be the tangent 
to OR at the point P, and the points P and R, be indefinitely near 
to each other, we have from the triangle RPS, 

dv 


(5) Cos.P —- 


(du? + dv’)? 
Equating (4) and (5), we have, 


MISCELLANEOUS PROBLEMS. 330 


1+ap+ be dv 
(6) Ser a at 
Matpag? Gira 
Also, the length of the element of the curve in space is the same 
as the length of the element of the curve in the development. 


Hence, we have, 
(7) dx? + dy? + dz? = dv* + du’, 

From the four equations, (2), (6), (7), eliminate 2,y,2, the result 
will be a differential equation between the co-ordinates w and », 
which when integrated, may be represented by 
(8) v = Qu. 

This is the equation of OR, the curve required. 

It is obvious that to eliminate a, y, and z, between (2), (6), (7), 
we must first by means of (2) integrate (7) for one of the co- 
ordinates x,y,z; and when these co-ordinates are eliminated we will 
have du and dv, involved in the resulting equation with the sign of 
integration. This sign of integration being disposed of by differen- 
tiation gives a differential equation of the second order whose in- 
tegral is equation (8). 


PROBLEM B. 

Given the equation of a plane curve, find the curve when. 
enveloped on a given cylinder. 

Let OP, fig. 96, be the plane curve, and (8) its equation. 

If the cylinder be placed with its element on OV, and rolled on 
the plane, the curve OR will be enveloped on the cylinder. Let the 
equation of the cylinder be 
(9) F(z,y,%) =.0. 

If now between the equations (6), (7), (8), we eliminate the co- 
ordinates u and », the result will be a differential equation which 
when integrated, we may represent by 
(10) Q(x,y,%) = 0. 

Between (9) and (10), eliminate first y and then 2, and we get 
(2), the equation of the enveloped curve. 

If the cylinder be perpendicular to the plane XY, then a and 6, 
in (1) are zero, and (6) is modified accordingly. 

7, 30 * 


¢ 
- 


334 INTEGRAL CALCULUS. 


If the given curve (8) be a straight line, the angle P is constant, 
and (6) becomes, 
1 + ap + bq 
MQ + p+ @)! 
This equation may be at once integrated, and we have (10). 
If a and b be zero, (11) becomes the partial differential equation, 
p+ ¢ = c, whose first integral is (612). 


(11) = C, a constant. 


PROBLEM C. 


A curve is traced ona given cone, determine the curve 
when the cone is developed on a plane. 


FIG. 97. 


Let (a',b',c',) be the vertex of the 
cone, The equation of the element of 
the cone is 
(12) z—a'=a(z—c'), y—b' = b(z—c’). 

Let the curve be, 

(13) oss Oz, Wf =r ¥2, 
and let the tangent to (13) be 
(14) L'— 2x = p(z —z), and y' —y = q(z' —2). 

Equation (4) expresses the cosine of the angle made by (12) 
with (14). As a and 6 in (12) vary with the position of the ele- 
ment, we must restore to (4) the value of M, and then eliminate a 
and 6 from (4), by means of (12), and we have, 

(15) Cos.P = F (2,y,2,p,q)- 

Suppose now the cone developed, and that CP is the curve in the 
development. Let the vertex of the cone be at A, then the element 
of the cone becomes the radius vector AP of the developed curve, 
the pole being the vertex. Then cos. P in the development, is [by 
Proposition D. 26, Appendix, Differential Calculus, ] 

dp 
(dp a p°dm*)? ’ 
and since the angle P is the same before and after development, we 
have by equations (15) and (16), 


(16) Cos,P = 


MISCELLANEOUS PROBLEMS. 335 


dp 
17 F Ys Ps ts + alts one} 
(17) (,Y57,P59) (dp? + p%da*)* 


Also the length of the element of the given curve being the same 
before and after development, we have, 

(18) dz? + dy? + dz? = dp’ + pda’. 

Eliminate x,y, and z, between the four equations (13), (17) and 
(18), and we have a differential equation, which when integrated, 
we may represent by 
(19) p = Pa. 

This is the developed curve CP. 

If the angle P be constant then (17) becomes, 


dp : 
(20) —— = 0, 
(dp? + p de’)* 
a constant, The integral of this gives us at once, for (19), the 
equation, 


Ca 


(21) Log. = a —c ian (. 


the logarithmic curve. 


PROBLEM D. 

Given the equation of a plane curve, find the curve when 
enveloped on a given conic surface. 

Taking for the conic surface the equation, 
(22) 9(x,y,2) = 0, 
and taking (19) for the given curve, we eliminate p and « between 
(17), (18), (19), and integrating the resulting equation, we have, 
(23) F (x,y,z) = 0. 

Eliminate first y and then a between (22) and (23), and we have 
(13), the equations of the enveloped curve. 


PROBLEM E. 


Determine the singular solution of a differential equation 
without first obtaining the complete primitive. 


336 INTEGRAL CALCULUS. 


In Proposition XV, Integral Calculus, we have shown the prin- 
ciple of singular solutions of differential equations, on the supposi- 
tion that we have the complete primitive of the equation. 

It may be convenient to deduce a method for obtaining such solu- 
tions when they exist without seeking first the integral of the 
equation 

Suppose we have the primitive equation, 

(24) H(2.9,c) = 0 =u, 

where c is the arbitrary constant, suppose that c enters into (24) in 
such a manner that it remains in the immediate differential of (24), 
then if 

(25) Mdzx + Ndy = 0, 

be the immediate differential of (24), we may get a differential 
equation by eliminating c between (24) and (25), Suppose the 
value of c from (25), be represented by 


(26) c= ?(%,Y,p). 
Put this value of ¢ into (24), and we have, 
(27) f (@9,0) =o =u’, 


where, for brevity, o is put for o(7,y,p). 

It is evident that equation (24) will be produced by integrating 
either (27) or (25). 

If (24) be solved for y, we may represent the result by 
(28) y =f (2c) 5 
and when this admits of a singular solution, we have, as in Propo- 
sition XV, 


(29) Sk? Sa, 


from which we deduce c = f(a), and eliminating c between (29) 
and (28), we get the singular solution sought. 

Let us now without obtaining its primitive (24), examine to what 
conditions the differential equation (27) is subject when it admits of 
a singular solution, For this purpose differentiate (27), and for 
convenience represent its differential [since @ contains x,y and p] by 
(30) du' = Pdx + Qdy + Rdp = 0, 
where P, Q, R, are functions of x,y, and p. Now regarding (28) 


MISCELLANEOUS PROBLEMS, 337 


as the integral of (27), we have y, a function of x and c, conse- 
quently p is also a function of 2 and c, and when (27) admits of a 
singular solution, c may be taken variable in (28), because in (29) 
we have c = fx. Hence, in order that (30) may be the complete 
differential of (27), we must put into (30) for dy and dp, their 
values in wt hee both of 2 and c. ‘These values are, 


(1) dy = BY ae + BA, and dp = SPAY 4 me 


ay et: f 


which being of into (30), te equation becomes, 
(32) du’ = (P i Solb aihth a) dx 4 (oe +R ) ac Rae 
dx de dc 


This is the differential of (27) for x,y and c variable. 

The term in (32) which is multiplied by dz, is evidently the same 
as equation (30) which being identically zero, independently of c, 
equation (32) reduces to 


(33) (a2 +R =) dag 
dc 
This is satisfied by making de = 9, or, 
dy dp 
34 Q— +R = 
(34) de 7p dc : 


The condition de = o gives c = a constant, which does not 
make known any thing concerning (27). The condition (34) being 
examined, we observe that in case (27) admits of a singular solu- 
tion, the condition (29) exists, which reduces (34) to 


(35) R s = 0; 

which is satisfied by putting, 

(36) sd tes D200. ky Sone, 
The condition that gave us (33) was, 

(37) du’ =P + pQ+RP=o 


which by virtue of the condition R = 0, becomes 
(38) P + pQ =) Op 


338 INTEGRAL CALCULUS. 


Now, though c does not enter expressly into (27), or its differ- 
ential (30), yet it is virtually contained in these equations, because 
it is the value of g which enters these equations, consequently, it 
may be regarded as existing implicitly in (80). The second con- 
dition of (36); viz. R = 0, must, therefore, exist in the case of a 
singular solution. This condition, and (38) may, therefore, be 
deduced immediately from (27). For if we solve (37) for the 
coefficient of R, we have, 


(39) apy ae kr tae 


which by conditions (36) and (38), becomes 


dp 0 

40 et ae 

i dx 0 
Hence in case (27) admits of a singular solution, (27) must be 


such that if we differentiate it for 2,y and p variable, and deduce 
from this differential the value of ee, we will have a fraction whose 
x 


numerator and denominator being each put equal to zero, will be the 
conditions of the singular solution. 

These conditions, viz. (38) and the last of (36), furnish two equa- 
tions which with each other, or with the given differential equation 
(27), will enable us to eliminate p. ‘The result, which may be re- 
presented by 
(41) o(7,y) = 9, 
will be the singular solution of (27), provided it satisfies that equa- 
tion. If the differential coefficient from (41) put into (27) does not 
render (27), when combined with (41), an identical equation, we 
conclude that (27) does not admit of a singular solution. 

In this procedure we have supposed the differential equation (27) 
freed from radicals, If that equation appears in the surd form, it 
should be first rationalised by squaring, d&c., before it is differ- 
entiated. 

We will give a few examples of the determination of singular 
solutions by this theory. 

Ex.1. Determine the singular solution of the equation 


MISCELLANEOUS PROBLEMS. 339 


(2) (c+ y)p—pe—(a+y) =o 
Differentiating this for x,y and p variable, we get 
iS dp_vt+p—p—P 
dx x+y — 2px 


The numerator of this is identically zero, and the denominator 
being put equal to zero, we have, 


(c) xrty—2pr=o. 
Eliminate p between (a) and (c), and we have, 
(d) (x — y) = 4ax. 


This satisfies the given differential equation (a). Hence, (d) is 
the singular solution of (a). 

Ex. 2. Finda curve whose normal varies as the part of the 
axis intercepted between it and the origin. 

By the enunciation we have, putting n for the ratio of the 
normal to the intercept, 
(e) yl +p’) = n(x + py)’. 

To determine whether this has a singular solution, we differen- 
tiate it as before, and putting numerator and denominator of DB 


br 
equal to zero, we find, 


(f) y = pr, 
and eliminating p between (f ) and (e), we find, 
ai Chine 
o V5 ae eee 
(8 J (1 abd n*)* 


This satisfies (e), and is, therefore, the singular solution. 

Ex. 3. Find the curve whose normal varies as the square root 
of the part of the axis intercepted between it and the origin. By 
the proposition, we have, 

(h) yl + p= n'(@ + py). 
Proceeding as before, we find for the singular solution, a parabola. 
Ex. 4. Given the equation. 
y = pe +(1 + p)*e. 
Prove that it has no singular solution. 
PROBLEM F. 
Determine the area of a plane curve. 


Proposition I. of the text exhibits the mode of determining the differential 
of a plane area by the method of indefinitely small quantities. ‘To deduce 
the differential of the same area by the method of limits proceed as follows ° 


4 Let GD be the curve and let the 
area sought be the part AGCB between 
the curve and the axis AX. 

Put the area AGCB = A, CB = y, 
AB = 2, BE=h. Complete the rec- 
tangles BR and BD. 

By Taylor’s Theorem we have for 
the ordinate DE the value 


| 340 INTEGRAL CALCULUS. 
t 


dy dy 5% 
DE dei veten J R24 &e. 
dx 2dx* 
The space DCBE is the increment of the area AGCB corresponding to the 
increment h of the abscissa. This space DCBE is contained between the 


rectangles BR and BD. The areas of these rectangles are 


BR eh sBD cane (y i afl cal gy 14h sc.) 
H 6 


d 2ax* 
DCBE is therefore contained between the values 
(m) hy and 
(2) ty ee ee B) 
dx Qd2z? 


But in the limit, the development 


dy DUS ve 

(°) Ay dz no 2dx° ete 
becomes simply y; and (mm) and (n) are equal, and the space DCBE becomes 
the differential of the area which is denoted by d.A. Hence the differential of 
the area being contained between the values (m) and (n) and these being in the 
limit equal, the differential of the area is equal to the value (m) and we have 
(p) d.A = yh = ydz, or integrating 
(7) A = f yd. 

Equations (p) and (q) are the same as found in Proposition I. of the text 
and are employed in the same manner as the equations of that Proposition. 

By similar reasoning we might proceed to determine by the method of limits, 
the properties of lines, surfaces, and solids which are presented in Proposition 
III. to XI. of the Integral Calculus. The results of the investigation would 
in all cases coincide with those presented in the text. Want of space however 
precludes us from giving more than the foregoing example of the method. 


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From Peter S. Duponceau, President of 
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PHILADELPHIA, Jan. 18th, 1844. 
GENTLEMEN,—I beg you will ac- 
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the young of 


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PHILADELPHIA, Jan. 20th, 1817 ; 
GENTLEMEN,—Having been request- 
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conviction, that the work of Fleming 


tria 
ts deh 


a 


and Tibbins is by far superior to its 
predecessors in the same line. Such 
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single letter A contains in Fleming and 
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a 


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I am very respectfully yours, 
F. A. BREGY. 


Boston, January 22, 1844. 

Ever since the first importation of 
Fleming and ‘Tibbins’ French and 
English Dictionary, I have constantly 
had it on my table, and have found it 
better than all other French diction- 
aries. I am therefore rejoiced to see 
an American abridged edition of so 
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which is most essential in the French 
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and valuable improvements are made. 
The mode in which the pronunciation 
is indicated is admirably plain and 
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Respectfully yours, 
GEORGE B. EMERSON. 


Boston Hicu ScHoon, 
February 1, 1843. § 

GENTLEMEN,—I have devoted some 
time and attention to the examination 
of Fleming and Tibbins’ French and 
English Dictionary, lately published 
by your firm ; and, although the merits 
of such a work can be thoroughly 
tested only by long use and a careful 
collation of it with kindred works, yet 
{ must say that this dictionary bears 
evident marks of its superiority to any 


other that has been introduced into this 


| country. 


By comparison, I find its vocabu- 
lary very copious and the idiomatic 
phrases quite numerous. The techni- 
cal terms are a very important addition, 
and the conjugation of verbs will prove 
of great use to the learner. The me- 
chanical execution of the work, which 
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a recommendation which immediately 
impresses itse]f on the eye. 

A complete and accurate dictiona- 
ry is of the utmost importance in the 
acquisition of a foreign language, and 
I feel justified in recommending this 
as one of great excellence 

Very respectfully, yours, 
THOMAS SHERWIN. 


From Isaac Leeser, Minister of the 
Hebrew Portuguese Congregation, 
Philadelphia. 

GENTLEMEN—It is with much plea- 
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During its progress through the press 
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convinced that it would prove an in- 
valuable aid to those who wish to 
acquire a knowledge of the most 
fashionable language of Europe. To 
its original and intrinsic merit is to be 
superadded the additions of the Ameri- 
can editor, who has enriched it with 
more than five thousand words (Medi- 
cal, Botanical, &c., &c.,) not in the 
French copy; also an excellent table 
of verbs, furnished by Mr. Picot. I 
cannot doubt that it will soon become 
an especial favourite with a discerning 
public; especially, as the moderate 
price you have fixed on it, a little more 
than one-fourth of the cost of the 
Paris edition, will bring this valuable 
Lexicon within the reach of the gene- 
ral student. Respectfully yours, 

Isaac LEESER 


PETER PARLEY’S 
COMMON SCHOOL HISTORY, 


WITH ONE HUNDRED AND FIFTY ENGRAVINGS, ILLUSTRATING 
HISTORY AND GEOGRAPHY. 


Price 75 cents. 


Tuts work is universally admitted to be the most successful attempt 
to bring General History within the scope of our schools and acade- 
mies, that has ever been made. The importance of having such a 
work in our seminaries, cannot be too highly estimated. Many chil- 
dren have no other means of education than are furnished by the pub- 
lic schools: if they do not here obtain the elements of Universal His- 
tory, they go through life in ignorance of a most important portion ot 
human knowledge. This work is calculated to remove the difficulties 
which have hitherto exchided this study from our schools. It presents 
Universal History in a series of interesting and striking scenes, weaving 
together an outline of Chronology, illustrated by descriptions, which, 
once impressed on the mind, will never leave it. One peculiar advan- 
tage of the work is, that History is here based upon geography, a 
point of the utmost importance. The success of the work, in actually 
interesting children in the study of history, has been practically tested 
and demonstrated. Several instances have occurred, in which pupils, 
before averse to history, have become deeply interested in it, preferring 
it to almost any other subject. The lessons are so arranged, that the 
whole study may be completed in a winter’s schooling. It is deemed 
particularly desirable that a subject so important should be introduced 
into all our common schools ; and as calculated to aid in such a pur- 
pose, the publishers invite the attention of all persons interested in 
education, to this work. 

It is only necessary to add, that this History has, in the short space 
of fifteen months, run through six large editions, and is now intro- 
duced into many of the best schools in the United States. It has been 
published in England, and has there met with the most decided 
favour. The following notices will show the estimate of it, formed by 
persons well qualified to decide upon its merits. 


RECOMMENDATIONS OF COMMON SCHOOL HISTORY. 


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ee 


RECOMMENDATIONS. 


From Professor C. D. Cleveland, Philadelphia. 


The publisher requests my opinion of Parley’s Common Schoo! History. It is seldom 
that I give an opinion upon school books, there are so few that I can recommend with 
a clear conscience ; and publishers do not wish, of course, to send forth a condemning 
sentence to the world. But in this case I can truly say that, having used the book in my 
school since it was published, I consider it a most interesting and luminous compend of 
general history for the younger classes of scholars; and that, were I deprived of it, I 
know not where I could find a work that I could use with so much pleasure to myself, 
and profit to those for whom it is designed. 

Respectfully yours, 
Cc. D. CLEVELAND. 


I fully concur in the above opinion of Professor Cleveland. 
JOHN FROST, 
Professor of Belles Lettres in the High Schoal of Philadelphia. 


From Rev. C. H. Alden, late Principal of the High School for Young 
Ladies, Philadelphia; now Chaplain in the U. S. Navy. 


I am greatly pleased with Parley’s Universal History, and shall introduce it into 
‘my school. Judicious in its arrangements, attractive in style, and striking in selec- 
tion, it commends itself to teachers and parents as particularly appropriate to the 
juvenile mind. When known to school committees, and others concerned with our 
eommon schools, it cannot fail of heing introduced into general use. The Geographical 
and Chronological accompaniments greatly enhance the value of the work; nor should 
the numerous and appropriate cuts and maps be overlooked, furnishing, as they do, a 
very valuable addition to both its beauty and utility. 


Philadelphia, September 19, 1839. 
Having examined Parley’s Common School History, I do not hesitate to say that, in 
my opinion, it is decidedly the best elementary general history I have seen, and I recom- 
mend its use to other teachers. 
M. L. HURLBUT. 


The above is concurred in by the undersigned, as follows: 
I intend to introduce it into the academical department of the University of Pennsyl- 


vania, under my care, as soon as possible. 
SAMUEL W. CRAWFORD. 


I have already introduced Parley’s Common School History as a class-book. 
SAMUEL JONES, 
Principal of Classical and Mathematical Institute. 


After a careful examination of the above-mentioned work, I am convinced that it is 
the best treatise for beginners in history, whether juvenile or adult, that I have ever seen. 
J. J. HITCHCOCK. 


I fully concur in the above. 


R. P. HUNT. 
I consider the work as a valuable acquisition in our schools for elementary classes in 
History. 
2 M. SEMPLE, M. D., 


Principal of Young Ladies’ Institute. 


Parley’s Common School History will, in my opinion, prove a valuable work in every 
school in which history is a subject of instruction. The modesty of its title is far from 
giving an adequate idea of its excellence. 

OLIVER A. SHAW. 


| 
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RECOMMENDATIONS OF COMMON SCHOOL HISTORY. 


The following resolution passed the Board of Controllers of the Public Schools of 
Philadelphia, September 9, 1841: 

“* Resolved, That the teachers of the different schools may, with the approval of the 
various Visiting Committees, establish, where the same has not already been done, His- 
tory classes; and that they use for the pupils in said classes, ‘ Parley’s Common School 
History,’ in addition to any other history heretofore authorized by the Board ;-~said book 
to be supplied at the cost of the respective pupils.’’ 

THOMAS B, FLORENCE, 

From the Minutes. Secretary. 


Philadelphia, April 30, 1842. 
Mr. E. H. BUTLER. 

Sir, -- I bave with much pleasure given to Parley’s Common School History a very 
thorough examination. Its object being to present to the youthful mind a brief and inte- 
resting view of the world’s history, as might be expected from the distinguished reputa- 
tion of its author, this end has been admirably attained. The most remarkable events 
which have occurred since the creation of the world, have been carefully selected ; and 
have been described in so perspicuous a manner, that even beginners in the study of his- 
tory can readily comprehend them. So felicitous are the descriptions, and such is the 
interest with which the author has succeeded in investing the subject, that the labour of 
study is converted into pleasure: and the impressions made, are not easily effaced. 

Yours truly, 
SAMUEL RANDALL. 
Young Ladies’ Seminary, 229 and 231 Arch street 


We fully concur in the above commendation of Parley’s Common School History, 
having used it with much satisfaction in our schools. 


P. A. CREGAR, E. ROBERTS, 

D. R. ASHTON, D. M. PADDACK, 
W. BRADFORD, C. PELTON, 
WILLIAM NOTSON, JOSEPH P. ENGLES, 
THOMAS T. AZPELL, THOMAS M’ADAM, 
SPENCER ROBERTS, JOHN EVANS, 
JOHN MOONEY, L. W. BURNET, 


Teacher, Fourth and Vine strects. 


Philadelphia, May 5, 1842. 
A careful examination of “ Parley’s Common Schoo! History,” gave me a very 
favourable opinion of its excellence; and the trial which I have since made of it as a 
class-book in my school, has convinced me that I did not overrate its merits. 
JOHN ALLEN, 
Principal of a Female Seminary, No. 274 North Seventh strect. 


I feei much interested in the volumes which were left for my examination. Parley’s 
Common School History is excellent in its way; the engravings serve to explain the 
dress, armour, &c. of former years to youth, who receive a more decided and satisfac- 
tory explanation by means of a cut, than by pages of description; the maps are equally 
appropriate. 

[ shall have no difficulty in forming a class of physiology, my pupils take hold of it 
with eagerness ; and treated so delicately, as it is in this book, I can see no objection to 
its introduction, while some knowledge of this aubject, amidst the onward progress of 
> re 4 *7 ¢ ~ ans ie 
education, will soon be regarded as indispensable W. CURRAN. 


Baltimore. 
Having examined Parley’s Common Schoo] History, 1 do not hesitate to say that, in 
my opinion, it is decidedly the best elementary Genera! History [ have seen, and | recom- 


and j or teachers. 
mend its use to other teachers A. B. CLEVELAND, 
Female Classical School. 


ee a en ES, FEN eee Sia he 


COMSTOCK’S ELOCUTION. 


A SYSTEM OF ELOCUTION, wir SPECIAL REFERENCE TO 
GESTURE, TO THE TREATMENT OF STAMMERING, AND DEFECTIVE 
ARTICULATION, COMPRISING NUMEROUS DIAGRAMS AND ENGRAVED 
FIGURES ILLUSTRATIVE OF THE SUBJECT, BY ANDREW CoMSTOCK, 
M.D., Princrpan oF THE VocaL and PotyeLoTr GYMNASIUM. 


ErenrH Epirion, ENLARGED. PHILADELPHIA: PUBLISHED BY 
E. H. Burier & Co. 


This is a duodecimo volume of nearly 400 pages. It is printed 
on good paper, and is well bound in leather. Price $1. 


( 


The work is theoretical and practical. It comprises a variety 
of exercises for the cure of defective articulation, cuts showing the 
best posture of the mouth in the energetic utterance of the ele- 
mentary sounds of the English language, numerous diagrams illus- 
trative of the melody and modulations of the voice, &c., and more 
than 200 engraved figures illustrative of the subject of gesture. 

Under the head, Pracricat Exocurion, are a variety of Kxer- 
cises on the Elements of the English Language, which are cal- 
culated to develope the voice, increase its compass, and give 
flexibility to the muscles of articulation. 

The Exercises in Reading and Declamation have been taken from 
some of the best ancient and modern authors; and they are well 
adapted to the purposes of the student in elocution. In these Ex- 
ercises, most of the sounds liable to be omitted, or to be imperfectly 
articulated, are represented by italic letters. Hence, the reader 
will have no difficulty in correcting all ordinary defects in his ar- 
ticulation. 

The arrangement of the several parts of the work is strictly 
systematic; each is discussed in its natural order, and with as much 
brevity as consists with perspicuity. ‘The work is designed for 
the use of colleges and schools, as well as for the instruction of 
private individuals who desire to improve themselves in the art of 
reading and speaking. 

The value of vocal gymnastics cannot be duly appreciated by 
those who have not experienced or witnessed their beneficial re- 
sults. But the time is not far distant when these exercises will 
be considered, by all intelligent persons, an essential part of 
primary instruction. 


STANDARD BOOKS PUBLISHED BY E. H. BUTLER AND CO. 


terete a Sr = 


From Benjamin Hallowell. Principal 
of the Friends’ High School, Phila- 
delphia. 

I have examined with some care the 

“ First Lines of Natural Philosophy,” 

by Reynell Coates, M.D., published 

by E. H. Butler & Co., and I believe it 
to be a work unusually well calculated 
to give to the student a clear under- 
standing of the different subjects upon 
it treads. The clear and fa- 
miliar style of the author, and his mode 
of preparing the learner to compre- 
hend the meaning of the different sci- 
entific terms, previous to their being 
used in the work, are advantages, in 
my opinion of no small merit. 

Signed, 
BENJAMIN HALLOWELL. 
Philadelphia, 11th mo. 20th, 1845. 


which 


From the Presbyterian. 


We have seldom been more gratified 
by the execution of an elementary 
book, than by the one before us. The 
intelligent reader will at once perceive 
that it is not a mere compilation, in 
which the materials are well selected 
and arranged, but the product of a 
philosophical mind—the result of much 
thought and skilful analysis. The 
author displays a clear comprehension 
of his subject, and he has succeeded 
remarkably in communicating an in- 
telligible view of it to his reader. In 
our opinion he has furnished one of 
the best text books we have met with, 
for the aid of those who wish to ob- 
tain a good general view of this in- 
teresting science. 


From the Alexandria (Va.) Gazette. 

This is a book of great merit. Dr. 
Coates is favourably known as the 
author of ‘‘ Physiology for Schools,” 
a work well received by Teachers, and 
now introduced extensively into the 
schools as a text book, in that branch 
of natural science. His work on 
Natural Philosophy is superior in 
respect to style and arrangement, 


to the popular work above mentioned. 
The book consists of about 400 pages, 
including an index and several pages 
of questions. The different branches 
of the science are discussed in a clear 
and foreible manner, avoiding the too 
frequent use of technical terms, and 
yet not altogether discarding them. 
This feature in the work, together with 
its intrinsic merit, renders it particu- 
larly valuable to the private learner. 
The majority of elementary books now 
in use, are composed of extracts taken 
verbatim from longer works, with here 
and there a few lines from the com- 
piler. Such books are always unsatis- 
factory to the inquiring mind of the 
student, and instead of producing in 
him ardour for further investigation, 
often disgust him with the study. Dr. 
Coates has not used the scissors.— 
Being intimately acquainted with the 
subject, he has condensed intoa medium 
sized book the valuable parts of the 
whole science, and, at the same time, 
preserves a free and attractive style. 
It would be out of place for us to ex- 
amine this work critically, as it would 
require more space than is usually 
allotted to notices of books; but we 
may recommend to the attention of 
those interested in the scientific pub- 
lications, the articles on Mechanics, 
Hydraulics, and Electro-Magnetism ; 
not that they contain anything par- 
ticularly new to those already versed 
in the science, but because they ex- 
hibit so rich an assemblage of useful 
facts rendered accessible at the same 
time by the pleasing manner of the 
writer. 


From the Philadelphia Ledger. 


In preparing this work the author 
seems to have had in view chiefly to 
render the instruction he imparts prac- 
tically useful to the student. It con- 
tains a number of familiar illustrations, 
which render the subject plain and 
easily to be comprehended, besides 


even | being interesting to the student. 


= 


= 4 


FIRST LINES OF PHYSIOLOGY. 


BEING AN INTRODUCTION TO THE SCIENCE OF LIFE. 


BY REYNELL COATES, M.D. 
Price $1 00. 


Tus work is designed expressly for the use of Schools, and has 
been carefully adapted to the capacities of Children, while the matter 
and style are such as to render it at once attractive and instructive to 
Youth who are advancing towards the conclusion of their studies, 
even in Seminaries of the highest class. 

The plan of the Work, and the mode in which the subject is 
handled, are in a great degree novel, and are the result of much 
thought on the proper manner of communicating scientific informa- 
tion to the young. The well-known literary character of the Author 
is a sufficient guaranty for the execution of this difficult task. 

Technical terms are avoided, as much as possible, and those which 
are necessarily employed, are fully explained in an accurate and 
simple manner. No term is given until the Student is impressed with 
the want of a word, to express an idea already received; so that the 
memory is not fatigued, at the very commencement of the study, with 
a long list of words, and abstract definitions, which he has no means 
of fixing in his mind by association. 

PuysloLocy is a subject of the deepest interest to all who are de- 
sirous properly to cultivate their powers of body and mind; and it is 
10w beginning to be conceded, that no course of education can be 
regarded as complete, without including some general knowledge of 
the science of Life. A Text-Book upon this subject is anxiously sought 
for by the leading teachers and professors of our country; but it has 
been supposed that, desirable as such knowledge must be for those 
who are charged with the care of the young, there is something in the 
nature of the study, rendering it unfit for introduction into seminaries 
for young ladies. ‘The error of this opinion is most clearly shown in 
the work now offered to the public. It contains not a word that can 
be regarded as objectionable by the most fastidious delicacy. 

The Author has treated the subject as though his readers were 
totally ignorant of living actions, and leads them forward from the 
simplest first principle, as deduced from the history of the lowest links 
of the chain of animated nature, to the consideration of the loftiest 
endowments of man: and so far is he from resorting to the easy ope 
ration of the scissors, in the manufacture of a book, that the entire 
volume does not contain an extracted sentence. An Appendix presents 
a complete Series of Questions as an assistance to Preceptors, referring 
by paragraphs to every fact and doctrine. 


RECOMMENDATIONS OF COATES ’s PHYSIOLOGY. 


RECOMMENDATIONS. 


1 have used Coates’s “ Physiology for Schools,’’ as a T'ext-Book on this subject, 
since its first publication, and think it superior to any work of the kind that has come 
under my notice. ‘he plan is excellent; at once scientific and popular—calculated, not 
to load the memory with barren facts, for the mere purpose of exhibition, but to interest 
the imagination, excite curiosity, and thus cherish habits of accurate observation and 
just reasoning. In the hand of a judicious teacher, it is capable of being made one of 
the most pleasant and useful of School Books. I should be gratified to see it in general 


use. 
° M. L. HURLBUT. 
I concur in the above opinion of Mr Hurlbut. 
JOHN FROST. 


Philadelphia, April 30, 1242. 
Mr. E. H. Bur er. 

Sir, — Please accept my thanks for a copy of the ** Physiology for Schools, by Rey- 
nell Coates, M.D.’’ It is unnecessary for me to speak of the value of the study; or of 
the general interest which has been awakened in its favour within a few years. The 
manner in which Dr. Coates has accomplished his object, is what particularly concerns 
us. With respect to this, I would briefly say ; if systematic arrangement ; if perspicuity 
of style; if clearness of illustration ; and if freedom from anything which can offend the 
most delicate and correct taste, be the standard of merit; then, it must be admitted, that 
the Author has done all that can be required. I trust you will receive from an en- 
lightened community, substantial proof, that it can appreciate what is truly useful. 

Yours, truly, 
SAMUEL RANDALL, 
Young Ladies’ Seminary, 229 and 231 Arch Street. 
May 5, 1842. 

| concur in the above expression of Samuel Randall, Esq., and trust that the commu- 

nity will encourage the use of Dr. Coates’s work. 
JOHN D. BRYANT, 
Principal of English and Classical Academy. 


I most heartily concur with the above recommendation. 
S. W. CRAWFORD. 


So far as we have had time to examine Dr. Coates’s Physiology, we agree with above 


recommendations. 
W. BRADFORD. L. W. BURNET, 
Teacher at corner 4th and Vine Streets. 


We, the undersigned, fully concur with the opinion expressed in the above recom- 
mendation. 


P. A. CREGAR, D. M. PADDACK, 

D. R. ASHTON, C. PELTEN, 

THOMAS BALDWIN, JAMES GOODFELLOW, 
WILLIAM NOTSON, J. WOOD, 

SPENCER ROBERTS, JOSEPH P. ENGLES, 
JOHN MOONEY, THOMAS M’ADAM, 

E. ROBERTS, JOHN EVANS, 


THOMAS T. AZPELL, 
Teacher, No. 35 Spruce Street. 


Philadelphia, May 7, 1842. 
With the plan and general arrangement of “ Coates’s Physiology,’’ I am much 
pleased. The principles of the science are presented in a familiar manner, calculated 
to interest as well as instruct the pupil. 


I shall introduce the work into my school. 
W. M. RICE 


HART’S ENGLISH GRAMMAR. 


An Exposition of the Principles and Usages of the English Language, by JOHN 
S. HART, A. M., Principal of the Philadelphia High School, and Member of the 


American Philosophical Society. 


Duodecimo. 


In the preparation of this work no 
special attempt has been made at no- 
velty. The author’s aim has been 
chiefly to make a careful and accurate 
digest of those principles of the language 
which have been remarked by pre- 
vious writers, and to state these princi- 
ples with precision and perspicuity. At 
the same time, the work is believed to 
contain many observations that are 
new, and a satisfactory solution of many 
difficulties not solved in other works on 
the subject. Unusual pains have been 
taken in preparing the parts intended to 
be committed to memory; viz., the 
Rules and Definitions. In respect 
to these, it is believed, the qualities of 
clearness, brevity, and accuracy have 
been attained to a high degree in the 
Grammar now offered to the public, 
Special care has been used also to distin- 
guish the parts which are elementary 
from those which are not. A great defect 
in the books heretofore used has been 
the want of sufficient care in this par- 
ticular. In some grammars, the whole 
text is of uniform size throughout. 
This makes it necessary either for the 
teacher to mark with a pencil the parts 
to be committed to memory, or for the 
pupil to burden his memory with the 
whole book, either of which is an into- 
lerable labour. Even in those works in 
which such a distinction is attempted, the 
parts in coarse print are generally en- 
tirely too numerous. The book is en- 
zumbered thereby, and the memory is 
burdened entirely beyond what is con- 
venient or proper. In the present Gram- 
mar nothing is put in large type but 
those rules and definitions which are 


Price 38 cents. 


of primary and indispensable import- 
ance. All the miscellaneous observa- 
tions, and all rules and definitions, of 
secondary importance, are printed ina 
type of medium size, clearly distin- 
guished to the eye from the others. 
There is also a third and large class of 
observations, partaking of the nature 
of discussions, and adapted to the wants 
of the teacher and of the advanced 
scholar rather than to the beginner. 
These portions are put in quite a 
small type. By this means a large 
amount of matter is brought within a 
small compass; and the character and 
relative importance of the various parts 
of the text are indicated clearly to the 
eye. At the bottom of each page are 
questions and exercises on all the mat- 
ter contained in that page. This gives 
great facility in hearing large classes, as 
it prevents the necessity of referring 
back and forth from the question to the 
answer. ‘The questions are separated 
from the text, being placed at the bot- 
tom of the page, so as not to embarrass 
those who do not wish to use them. 
They are so prepared as to facilitate 
the despatch of lessons to those teachers 
who have large and numerous classes, 
and at the same time they are of a sug- 
gestive character, and suited to the 
wants of those who wish to vary the 
exercises from time to time. The dif- 
ferent parts of the subject are clearly 
divided and classified under distinct 
heads. But the paragraphs are all 
numbered consecutively from begin- 
ning to end. This gives great facility 
in making references. 


STANDARD BOOKS PUBLISHED BY E. H. BUTLER AND CO. 


i ie RE OS... ee a earns eR LT 
useful information; and a large collection of verbal distinctions with 
‘liustrations. Designed as a Reading book for the highest classes in 
Academies and Schools. Price 75 cts. 

The whole forming a Series of interesting, useful, and economical 
School Books. 

They have been through several large editions, and the Publishers 
have had the whole Series stereotyped, so that future editions will be 
uniform, in every respect. They have spared neither pains nor ex- 
pense to render these School Books worthy of attention, and they 
fatter themselves that their appeals to an enlightened public, for 
encouragement, will not be made in vain. 

These Six Volumes, compiled by OL1vER ANGELL, comprise a Series 
vhich is nndoubtedly more suitable for the purpose for which they 
are designed, than any previous publications ; and they are more 
popular among those who have the direction of education than any 
ever prepared in this country. There is a decided advantage in pos- ~ 
sessing sets of Elementary books by the same author, who has pursued 
a similar plan with each, rising step by step, and who, it is presumed, 
would be better able to preserve the proper gradation of style and 
matter, than several individuals would. ‘This will be found to be the 
case in the present Series — and no matter which appears in one, is 
inserted in either of the others; so that the six compilations may be, 
and are intended to be used, in different classes in the same school. 


RECOMMENDATIONS. 


Columbia, Pa., December 8, 1840. 
To THE TEACHERS IN THE Pugtic ScHOOLS. 

The bearer is authorized, by the Board of Directors, to visit the Schools, and make 
arrangements for the adoption of ‘* Angell’s Series of Reading Books ;’’ * Parley’s 
History ;’’ and ‘* Smith’s Grammar,’”’ in all our Schools. It is the intention of the 
Board to insist upon each scholar’s providing himself, or herself, with such of the above 
works as the Teacher thinks necessary. You will, therefore, assist by giving him all 
the information he desires. 


By authority of the Board of Directors. 
JOHN F. HOUSTON. 


—_—_—— 


Philadelphia, January 9. 

You wil! accept my sincere acknowledgments for the set of Angell’s Class Books, 
which you were so obliging as to send me a few weeks ago. 

This Series, embracing a range of exercises from the easiest prose composition to the 
highest efforts of description, poetry, and eloquence, appears likely, in the hands of dis- 
creet and able teachers, to lead the youthful understanding and taste onward from ene 
degree of development to another, while the moral sentiments and social affections are 
duly fostered and improved. I cannot doubt, that in Common Schools, this Series 0 
Reading Books will find a favourable reception, while some of the superior numbers 
will be able to sustain a creditable competition with works of a similar rank already in 
use in our Academies and High Schools. 

Very respectfully yours, Ww. R. JOHNSON, 
Professor of Mathematics and Moral Philosophy 


Gh ta lh 5 od Wl eo 


OR, 


A DESCRIPTION OF THE STARRY HEAVENS. 


Designed for the use of Schools and Academies; accompanied by an Atlas of 


the Heavens, showing the places of the principal stars, clusters, and nebulz. 
By E. OTIS KENDALL, Professor of Mathematics and Astronomy in the 


Central High School of Philadelphia, and Meinber of the American Philosoph- 


ical Society. The Uranography contains 365 pages 12mo, with 9 fine engray- 


ings. 
raphy and Atlas $1.25. 


URANOGRAPHY, as the word imports, 
is simply a description of the Heavens. 
It is Descriptive, as distinguished from 
Practical Astronomy. Astronomy, as 
a practical science, requires the use of 
costly instruments and a knowledge of 
the higher mathematics. But Uranog- 
raphy requires for its study no more 
expensive apparatus or higher attain- 
ments than Geography. The same 
boy or girl who is competent to study 
the description of the earth, may with 
equal ease, and from the same teacher, 
learn a description of the heavens. 
The ‘ Uranography and Atlas” are to 
the one study what the “ Geography 
and Atlas’? are to the other. 

A great obstacle to the study of Uran- 
ography heretofore, has been the diffi- 
culty of transferring to the heavens 
themselves, the ideas acquired in study- 
ing the maps. There was so much 
in the map that was not in the heavens, 
that it was extremely difficult for a be- 
ginner to conceive the one to be in any 
respect the representative of the other. 
A Celestial Map or Globe, crowded 
with highly coloured pictures of birds, 
and beasts, and four-footed animals, 


The Atlas is in 4to, and contains 18 large maps. 


Price of the Uranog- 


and creeping things, might well look, 
to the eye of the uninitiated, more like 
the show-bill of a menagerie than a 
picture of the starry heavens. In the 
present work, however, while a faint 
outline of the old constellations is pre- 
served for the sake of their historical 
associations, prominence is given in the 
maps to that which is prominent in 
the heavens, viz: to the sTARS THEM- 
SELVEs. 

This feature of the work is made yet 
more striking by the introduction ot 
another, of a character altogether no- 
vel. Not only are the objects which 
are not seen in the heavens, excluded 
from the maps, but the heavens them- 
selves are represented more nearly in 
their true colors. Instead of painting, 
as heretofore, the stars black and the 
sky white, the groundwork of the map 
is here the deep blue of heavenly space, 
while the stars are a brilliant, spotless 
white. Such a representation of the 
subject seems to be the one least like- 
ly to confuse the mind of the student, 
when, from a contemplation of his Ce- 
lestial Atlas, he turns to contemplate 
the august scene which it represents. 


STANDARD BOOKS PUBLISHE 


Uranocrapny.—‘‘ I have often won- 
dered that the mighty Heavens, the 
richest page in all the works of God 
which we gaze upon, are not made the 
subject of study in our higher schools. 
To be sure, we here and there find a 
celestial globe in the corner of the 
school room; but we seldom find the 
little scholar who can go out and select 
from the glowing heavens the different 
classifications of the constellations. I 
mourn that it is so. And I rejoice to 
have a book which ought to bea favour- 
ite through the land. This work, with 
its eighteen maps, is abundant, in the 
hands of a competent teacher, to create 
pure pleasures which shall be renewed 
and deepened through life.”—Rev. J. 
Todd, D.D. 


Uranocrapuy.—“* A capital work 
for schools and academies.?’—Richmond 
Inquirer. 


Uranocrapny.—* No text book on 
the subject has ever been published at 
all equal to this of Prof. Kendall’s.”— 
Watchman of the South. 


—— 


Uranocrapuy.—‘* The author has | 


brought to the work a strong love for, 
which has led to a deep knowledge of, 
the science of which he treats; and 
his position as a teacher, enables him 
to set forth his instruction in an avail- 
able form. We cordially recommend 
the work to parents and teachers, as 
eminently calculated to assist the young 


in arriving at a knowledge of the things 
‘that are in the Heavens above.’”—U 


S. Gazette. 


Uranocrapuy.—* We commend this 
work with warmth and confidence. It 
is admirable in its arrangement and de- 
tails—is clear, comprehensive, and in- 
structive.”—Philada. Inquirer. 


D BY E. H. BUTLER AND CO. 
DS ee 
Uranocrapuy.—* With the author 
of this work Astronomy may fairly be 
He is one of the most 
enthusiastic pursuers of the science in 


styled a passion. 


the world; and as a consequence, he is 
one of the most competexrt men to treat 
of it, so far as knowledge is concerned 
His profession has given him what all 
do not possess—the capacity to impart 
information ; and we have in this work 
one most excellent to follow the ordi- 
nary elementary treatises.’’—Saturday 
Post. 


UranoGrapuy. — ‘It appears to us 
that this work supplies a desideratuin 
with the schools, and will much facili- 
tate the study of the ‘ wonders of the 
Heavens.2 We commend it to the at- 
tention of teachers, and doubt not they 
will find it a valuable acquisition to 
their means of imparting knowledge.— 
Richmond Compiler. 


——— 


Uranocrapuy.— The arrangement 
and details of this work are excellent. 
The author appears to have selected for 
elucidation the prominent points in As- 
tronomy, and has explained them with 
clearness and simplicity. His remarks 
on Double Stars and Nebule go more 
into the profundity of the science, and 


lare able and interesting. We are glad 


to see practical teachers turning their 
attention to the preparation of school 
books. There is no other way in which 
they can employ their leisuze hours 
with so much advantage to their pupils 
and the community—Baltimore Clip- 


per. 


Uranograpuy.—‘ We know of no 
work of the same nature equal to this 
}in comprehensiveness and arrangement. 
It is designed for academies and schools, 
and will no doubt be generally adopt- 


|ed.2°—North American. 


FIRST LINES OF NATURAL PHILOSOPHY, 


Divested of Mathematical formule: being a practical and lucid introduction to 


the study of the science. Designed for the use of Schools and Academies, 


a 


and for readers generally, who have not been trained to the study of the 


exact sciences, and for those who wish to enter understandingly upon the study 
of the exact sciences. By REYNELL COATES, M. D., author of Physiology 


for Schools. Illustrated by 264 cuts. 

This is another most valuable con- 
tribution to the science of elementary 
instruction, from the pen of a writer 
already widely known as a lucid rea- 
soner, anda peculiarly happy instructor 
of youth. 

The mode in which the subject is 
discussed resembles that which has 
secured such general approval, and ex- 
tensive patronage for the ‘‘ Physiology 
Unlike most works de- 

similar purpose, this 


for Schools.” 
signed for a 
volume is not a compilation merely. 
The author has evidently considered 
the capacities and tastes of his audience, 
matured his plan, and mastered all the 
necessary relations of his theme before 
putting pen to paper; then, with the 
whole subject before him, and con- 
sidering his pupil as utterly ignorant of 
the first principles of nature, he begins 
as though addressing the extremely 
young; and, throughout the entire 
work, he nowhere oversteps the ability 
of the pupil—not a technical term is 
used unless fully and clearly explained, 
and no previous mathematical know- 
ledge is demanded of the student; al- 
though the reader is drawn, by light 
and easy, but logical and orderly stages, 
from the consideration of the simplest 
accidents of every day life, to the com- 
prehension of some of the grandest 
phenomena connected with Astronomy. 

Originality to any great extent would 
be out of place in the ‘ First Lines” 


Price 75 cents. 


of a science which has been cultivated 
for many ages; but in the sections on 
‘‘ Extension,’?? much novelty will be 
discovered, at least in the deductions, 
if not in the facts; while in the depart- 
ment of the Phenomena of Fluids, 
the text is occasionally enlivened by 
the valuable results of personal ob- 
servation. 

It is one of the peculiarities of this 
writer, that whatever he attempts to 
teach he teaches thoroughly; and 
while the Appendix of well-digested 
questions and references, greatly in- 
creases the value of the work to the 
practical preceptor of children acqui- 
ring the rudiments of education, its 
text will render it scarcely less valu- 
able to the youth about commencing 
the study of the exact sciences, the man 
of liberal information, and even the 
professional teacher. 

The study of the elements of natu- 
ral philosophy is now pursued in every 
well-ordered school or academy, and 
we believe that this volume will bear 
favourable comparison with any treatise 
of similar intention now offered to the 
public. 


CENTRAL HicH ScHOOL, : 
PHILADELPHIA, Sep?. 1846. 


Coates’ First Lines of Natural Phi- 
losophy, is used as a text book Yor the 
junior classes in this institution. 

Joun 8. Hart, Principal. 


NUGENTS’ 
FRENCH AND ENGLISH DICTIONARY. 


A Pocket Dictionary of the French and English Languages. 


In two parts. 


Price 63 Cents. 


1. French and English.—2. English and French. Containing a] 


the words in general use, and authorized by the best writers. The several 


parts of speech—The genders of the French nouns—The accents of the Eng- 


lish words, for the use of foreigners—An alphabetical list of the most usuitt 
Christian and Proper Names ; and of the most remarkable places in the know 
world—By THOMAS NUGENT, LL.D. The fifth American, from the las! 


London edition ; 


and Le Clerc’s last edition of the National French Dictionary ; 


with the addition of the new words, inserted in Moutardier’s 


the irregulari- 


ties of the English verbs and nouns ; and a comprehensive view of the pronun- 


ciation and syntax of the French language. 


By J. OUISEAU, A. M. 


FLEMING AND TIBBINS’ 
FRENCH AND ENGLISH DICTION ARY, 


ABRIDGED AND ADAPTED TO THE USE OF 


ACADEMIES AND SCHOOLS: 


By CHARLES PICOT, Esq., Professor of French in the University of Penn- 
sylvania, and JUDAH DOBSON, Esq., Member of the American Philosophi- 
eal Society, of the Academy of Natural Sciences, &c. &e. 


724 Pages 12mo. 


THE first edition in royal octavo of 
the French and English and English 
and French Dictionary, on the basis of 
that of Professors Fleming and Tib- 
bins, published in 1844, has been pro- 
nounced by many competent judges as 
the best and at the same time the cheap- 
est publication of the kind. The French 
scholar, the literary man, the physi- 
cian, the savant, in general, hold it in 
high estimation, and indeed prefer it to 
any other, on account of its compre- 
hensiveness and accuracy. But 
though the price of this library edition 
is extraordinarily low, a regret has 
often been expressed by those acquain- 
ted with its merits, that it should not 
be within the reach of every one: for 
this reason the editor has been induced 
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which will be found all the words in 
general use, a literal and figured pro- 
nunciation, together with fall instruc- 


al- 


tions adapted for the use of the English | 


Price $1.25. 

student of the French language. 
sides this, the tables of the verbs by 
Mr. Picot have been added, as being 
calculated to facilitate the study of this 
difficult part of the French language. 
In these tables—it will be seen—the 


Be- 


verbs are numbered, and so arranged 
as to show, ata glance, the formation 
of the various tenses—simple and com- 
pound ; the irregularities ; and modes 
of conjugation — affirmatively, nega- 
tively, and interrogatively. To the 
different verbs, as they occur in the 
body of the Dictionary, a number is af- 
fixed referring to the tables; and as 
their pronunciation is distinetly indi- 
cated, the work may be considered as 
affording a complete and ready means 
of ascertaining the modes of conjuga- 
tion, and the pronunciation of the verbs 
of the French language in all their 
forms—a desideratum not to be found 
in any other publication of the same 
nature. 
PHILADELPHIA, January, 1845 


WALKER’S PRONOUNCING DICTIONARY. 


Octavo. 782 pages. 


Price :—Fine edition $2.50—Common edition $1.25, 


& Critical Pronouncing Dictionary, and Expositor of the English Language ; 


which not only the meaning of every word is explained, and the sound of 
every syllable distinctly shown, but where words are subject to different pro- 
nunciation, the authorities of our best pronouncing dictionaries are fully ex- 
hibited, the reasons for each are at large displayed, and the preferable pro 
nunciations pointed out. To which is prefixed Principles of English Pronun- 
ciation, in which the sounds of letters, syllables, and words are critically 
investigated and systematically arranged; the influence of the Greek and 
Latin accent and quantity on the accent and quantity of the English, is 
thoroughly examined and clearly defined ; and the analogies of the language 
are so fully shown as to lay the foundation of a consistent and rational pro 
nunciation. Likewise, Rules to be observed by the natives of Scotland, Ire 
land, and London, for avoiding their respective peculiarities, and Directions t 
Foreigmers for acquiring a knowledge of the use of this Dictionary. The 
whole interspersed with observations, Etymological, Critical, and Grammat- 
ical. To which is annexed a Key to the Classical Pronunciation of Greek, 
Latin, and Scripture Proper Names, etc. By JOHN WALKER, Author of 


Elements of Elocution, Rhyming Dictionary, etc. 


In offering to the public a new Edi- 
tion of WaLkER’s DicrionaRy, the 
Publishers do not feel it to be necessary 
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merits of the Work itself. It is believed 
io be regarded very generally, on both 


lated, and their mistakes carefully cor- 
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are such as almost to stagger belief. 
These errors appear to have been creep- 
ing in for the last half century, each 


sides of the Atlantic, as the best and 
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guage, for the purpose of immediate 
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For this purpose, the existing editions, 
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edition repeating the mistakes of its 
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aim of the Publishers, in the present 
Edition, by a careful revision of pre- 


| vious ones, and by a diligent use of the 


other necessary means, to offer a text 


} as nearly as possible FAULTLESS 


SMITH’S 
PRODUCTIVE GRAMMAR. 


EneLish GRAMMAR ON THE Propuctive System: A method 
of instruction recently adopted in Germany and Switzerland. 
Designed for Schools and Academies. By Roswell C. Smith, 
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“The above work was composed, as is indicated by the title, on 
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have prevailed there, may not be irrelevant, on the present occasion, 
as they assist in forming an opinion of the comparative merits of the 
‘Productive system,’ on which this work is principally based,” &c. 
&c, &c. (Vide Preface of the work.) 

This work has been before the public several years, and its merits 
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for it has steadily increased; and it now enjoys the approbation of 
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his possession hundreds of Recommendations, from Teachers, School- 
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Copies furnished teachers, for examination, gratis. 


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